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    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

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    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

    Edited By Ramraj Saini | Updated on Oct 09, 2023 05:20 PM IST

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

    Algebraic Expressions and Identities Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by experts team at Careers360 team considering the latest syllabus and pattern of CBSE 2023-24. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques. Important topics like the product of the equation, finding the coefficient of the variable in the equation, subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter. Also Practice NCERT solutions for class 8 maths to command the concepts.

    Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

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    Algebraic Expressions and Identities Class 8 Solutions - Important Formulae

    • (a + b)2 = a2 + 2ab + b2

    • (a - b)2 = a2 - 2ab + b2

    • (a + b)(a - b) = a2 - b2

    • (x + a)(x + b) = x2 + (a + b)x + ab

    • (x + a)(x - b) = x2 + (a - b)x - ab

    • (x - a)(x + b) = x2 + (b - a)x - ab

    • (x - a)(x - b) = x2 - (a + b)x + ab

    • (a + b)3 = a3 + b3 + 3ab(a + b)

    • (a - b)3 = a3 - b3 - 3ab(a - b)

    Free download NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities for CBSE Exam.

    Algebraic Expressions and Identities Class 8 NCERT Solutions (Intext Questions and Exercise)

    NCERT Solutions to Exercises of Chapter 9: Algebraic Expressions and Identities

    what are expressions?

    Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

    Answer:

    Five examples of expressions containing one variable are:

    x^{^{4}}, y, 3z, p^{^{2}}, -2q^{3}

    Five examples of expressions containing two variables are:

    x + y, 3p-4q,ab,uv^{2},-z^{2}+x^{3}

    Question: 2(i) Show on the number line

    x

    Answer:

    x on the number line:

    1643105164197

    Question: 2(ii) Show on the number line :

    x-4

    Answer:

    x-4 on the number line:

    1643105231337

    Question: 2(iii) Show on the number line :

    2x+1

    Answer:

    2x+1 on the number line:

    c360_4-1


    Question: 2(iv) Show on the number line:

    3x-2

    Answer:

    3x - 2 on the number line

    1643105272368

    Algebraic expressions and identities class 8 solutions - Topic 9.2 Terms, Factors and Coefficients

    Question:1 Identify the coefficient of each term in the expression.

    x^2y^2-10x^2y+5xy^2-20

    Answer:

    coefficient of each term are given below

    \\The\ coefficient\ of\ x^{2}y^{2}\ is \1\\ \\The\ coefficient\ of\ x^{2}y\ is \ -10\\ \\The\ coefficient\ of\ xy^{2}\ is \5\\

    Algebraic expressions and identities class 8 ncert solutions - Topic 9.3 Monomials, Binomials and Polynomials

    Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.

    -z+5

    Answer:

    Binomial since there are two terms with non zero coefficients.

    Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.

    x+y+z

    Answer:

    Trinomial since there are three terms with non zero coefficients.

    Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.

    y+z+100

    Answer:

    Trinomial since there are three terms with non zero coefficients.

    Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.

    ab-ac

    Answer:

    Binomial since there are two terms with non zero coefficients.

    Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.

    17

    Answer:

    Monomial since there is only one term.

    Question: 2(a) Construct 3 binomials with only x as a variable;

    Answer:

    Three binomials with the only x as a variable are:

    \\ \\x+2,\ x +x^{2},\ 3x^{3}-5x^{4}

    Question: 2(b) Construct 3 binomials with x and y as variables;

    Answer:

    Three binomials with x and y as variables are:

    \\ \\x+y,\ x-7y, xy^{2} + 2xy

    Question: 2(c) Construct 3 monomials with x and y as variables;

    Answer:

    Three monomials with x and y as variables are

    \\ xy,\ 3xy^{4},\ -2x^{3}y^{2}

    Question: 2(d) Construct 2 polynomials with 4 or more terms .

    Answer:

    Two polynomials with 4 or more terms are:

    a+b+c+d, x-3xy+2y+4xy^{2}

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.4 Like and Unlike Terms

    Question:(i) Write two terms which are like

    7xy

    Answer:

    \\Two\ terms\ like\ 7xy\ are:\\ -3xy\ and\ 5xy

    Question:(ii) Write two terms which are like

    4mn^2

    Answer:

    \\Two\ terms\ which\ are\ like\ 4mn^{2}\ are:\\ mn^{2}\ and -3mn^{2.}

    we can write more like terms

    Question:(iii) Write two terms which are like

    2l

    Answer:

    \\Two\ terms\ which\ are\ like\ 2l\ are:\\ l\ and\ -3l

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Exercise: 9.1

    Question:1(i) Identify the terms, their coefficients for each of the following expressions.

    5xyz^2-3zy

    Answer

    following are the terms and coefficient

    The terms are 5xyz^{2}\ and\ -3zy and the coefficients are 5 and -3.

    Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

    1+x+x^2

    Answer:

    the following is the solution

    \\The\ terms\ are\ 1,\ x,\ and\ x^{2}\ and\ the\ coefficients\ are\ 1,\ 1,\ and\ 1\ respectively.

    Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

    3-pq+qr-rp

    Answer:

    The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

    Question:1(v) Identify the terms, their coefficients for each of the following expressions.

    \frac{x}{2}+\frac{y}{2}-xy

    Answer:

    \\The\ terms\ are\ \frac{x}{2},\ \frac{y}{2}\ and\ -xy\ and\ the\ coefficients\ are\ \frac{1}{2},\ \frac{1}{2}\ and\ -1\ respectively.

    Above are the terms and coefficients

    Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

    0.3a-0.6ab+0.5b

    Answer:

    The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

    Question: 3(i) Add the following.

    ab-bc , bc -ca, ca-ab

    Answer:

    ab-bc+bc-ca+ca-ab=0.

    Question:3 (ii) Add the following.

    a-b+ab, b-c+bc, c-a+ac

    Answer:

    \\a-b+ab+b-c+bc+c-a+ac\\ =(a-a)+(b-b)+(c-c)+ab+bc+ac\\ =ab+bc+ca

    Question:3 (iii) Add the following

    2p^2q^2-3pq+4, 5+7pq-3p^2q^2

    Answer:

    \\2p^{2}q^{2}-3pq+4+5+7pq-3p^{2}q^{2}\\ =(2-3)p^{2}q^{2} +(-3+7)pq +4+5\\ =-p^{2}q^{2}+4pq+9

    Question: 3(iv) Add the following.

    l^2+m^2+n^2 , n^2+l^2, 2lm+2mn+2nl

    Answer:

    \\l^{2}+m^{2}+n^{2}+n^{2}+l^{2}+2lm+2mn+2nl\\ =2l^{2}+m^{2}+2n^{2}+2lm+2mn+2nl

    Question: 4(a) Subtract 4a-7ab+3b+12 from 12a-9ab+5b-3

    Answer:

    12a-9ab+5b-3-(4a-7ab+3b+12)
    =(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
    =8a-2ab+2b-15

    Question: 4(b) Subtract 3xy+5yz-7zx from 5xy-2yz-2zx+10xyz

    Answer:

    \\5xy-2yz-2zx+10xyz-(3xy+5yz-7zx)\\ =(5-3)xy+(-2-5)yz+(-2+7)zx+10xyz\\ =2xy-7yz+5zx+10xyz

    Question: 4(c) Subtract 4p^2q - 3pq + 5pq^2 - 8p + 7q - 10 from 18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q

    Answer:

    \\18-3p-11q+5pq-2pq^{2}+5p^{2}q-(4p^{2}q-3pq+5pq^{2}-8p+7q-10)\\ =18-(-10)-3p-(-8p)-11q-7q+5pq-(-3pq)-2pq^{2}-5pq^{2}+5p^{2}q-4p^{2}q\\ =28+5p-18q+8pq-7pq^{2}+p^{2}q

    NCERT class 8 maths chapter 9 question answer - Topic 9.7.2 Multiplying Three or More Monomials

    Question:1 Find 4x\times 5y\times 7z . First find 4x\times 5y and multiply it by 7z ; or first find 5y \times 7z and multiply it by 4x .

    Answer:

    \\4x\times 5y\times 7z\\ =(4x\times 5y)\times 7z\\ =20xy\times 7z\\ =140xyz\\ \\4x\times 5y\times 7z\\ =(5y\times 7z)\times 4x\\ =35yz\times 4x\\ =140xyz

    We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

    Class 8 maths chapter 9 question answer - exercise: 9.2

    Question: 1(i) Find the product of the following pairs of monomials.

    4,7p

    Answer:

    4\times 7p=28p

    Question:3 Complete the table of products.


    First monomial \rightarrow

    Second monomial \downarrow

    2x

    -5y

    3x^2

    -4xy

    7x^2y

    -9x^2y^2

    2x

    4x^2


    ...

    ...

    ...

    ...

    ...

    -5y

    ...

    ...

    -15x^2y

    ...

    ...

    ...

    3x^2

    ...

    ...

    ...

    ...

    ...

    ...

    -4xy

    ...

    ...

    ...

    ...

    ...

    ...

    7x^2y

    ...

    ...

    ...

    ...

    ...

    ...

    -9x^2y^2

    ...

    ...

    ...

    ...

    ...

    ...

    Answer:

    First monomial \rightarrow

    Second monomial \downarrow

    2x

    -5y

    3x^{2}

    -4xy

    7x^{2}y

    -9x^{2}y^{2}

    2x

    4x^{2}

    -10xy

    6x^{3}

    -8x^{2}y

    14x^{3}y

    -18x^{3}y^{2}

    -5y

    -10xy

    25y^{2}

    -15x^{2}y

    20xy^{2}

    -35x^{2}y^{2}

    45x^{2}y^{3}

    3x^{2}

    6x^{3}

    -15x^{2}y^{}

    9x^{4}

    -12x^{3}y

    21x^{4}y

    -27x^{4}y^{2}

    -4xy

    -8x^{2}y

    20xy^{2}

    -12x^{3}y

    16x^{2}y^{2}

    -28x^{3}y

    36x^{3}y^{3}

    7x^{2}y

    14x^{3}y

    -35x^{2}y^{2}

    21x^{4}y

    -28x^{3}y^{2}

    49x^{4}y^{2}

    -63x^{4}y^{3}

    -9x^{2}y^{2}

    -18x^{3}y^{2}

    45x^{2}y^{3}

    -27x^{4}y^{2}

    36x^{3}y^{3}

    -63x^{4}y^{3}

    81x^{4}y^{4}

    Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

    2p,4q,8r

    Answer:

    the volume of rectangular boxes with the following length, breadth and height is

    \\Volume=length\times breadth\times height\\ =2p\times 4q\times 8r\\ =8pq\times 8r\\ =64pqr

    Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

    xy, 2x^2y, 2xy^2

    Answer:

    the volume of rectangular boxes with the following length, breadth and height is

    \\Volume=length\times breadth\times height\\ =xy\times 2x^{2}y\times 2xy^{2}\\ =2x^{3}y^{2}\times 2xy^{2}\\ =4x^{4}y^{4}

    Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

    a, 2b, 3c

    Answer:

    the volume of rectangular boxes with the following length, breadth and height is

    \\Volume=length\times breadth\times height\\ =a\times 2b\times 3c\\ =2ab\times 3c\\ =6abc

    Question:5(i) Obtain the product of

    xy,yz,zx

    Answer:

    the product

    \\xy\times yz\times zx\\ =xy^{2}z\times zx\\ =x^{2}y^{2}z^{2}

    Question:5(ii) Obtain the product of

    a,-a^2,a^3

    Answer:

    the product

    \\a\times (-a^{2})\times a^{3}\\ =-a^{^{3}}\times a^{3} =-a^{6}

    Question:5(iii) Obtain the product of

    2,\ 4y,\ 8y^{2},\ 16y^{3}

    Answer:

    the product

    \\2\times 4y\times 8y^{2}\times 16y^{3}\\ =8y\times 8y^{2}\times 16y^{3}\\ =64y^{3}\times 16y^{3}\\ =1024y^{6}

    Question:5(iv) Obtain the product of

    a, 2b, 3c, 6abc

    Answer:

    the product

    \\a\times 2b\times 3c\times 6abc\\ =2ab\times 3c\times 6abc\\ =6abc\times 6abc\\ =36a^{2}b^{2}c^{2}

    Question:5(v) Obtain the product of

    m, -mn, mnp

    Answer:

    the product

    \\m\times (-mn)\times mnp\\ =-m^{2}n\times mnp\\ =-m^{3}n^{2}p

    Class 8 maths chapter 9 NCERT solutions - Topic 9.8.1 Multiplying a Monomial by a Binomial

    Question:(i) Find the product

    2x(3x+5xy)

    Answer:

    Using distributive law,

    2x(3x + 5xy) = 6x^2 + 10x^2y

    Question:(ii) Find the product

    a^2(2ab-5c)

    Answer:

    Using distributive law,

    We have : a^2(2ab-5c) = 2a^3b - 5a^2c

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topic 9.8.2 Multiplying A Monomial By A Trinomial

    Question:1 Find the product:

    (4p^2+5p+7)\times 3p

    Answer:

    By using distributive law,

    (4p^2+5p+7)\times 3p = 12p^3 + 15p^2 + 21p

    Class 8 maths chapter 9 NCERT solutions - exercise: 9.3

    Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.

    4p, q+r

    Answer:

    Multiplication of the given expression gives :

    By distributive law,

    (4p)(q+r) = 4pq + 4pr

    Question:1(ii) Carry out the multiplication of the expressions in each of the following pairs.

    ab, a-b

    Answer:

    We have ab, (a-b).

    Using distributive law we get,

    ab(a-b) = a^2b - ab^2

    Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.

    a+b, 7a^2b^2

    Answer:

    Using distributive law we can obtain multiplication of given expression:

    (a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3

    Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.

    a^2-9,4a

    Answer:

    We will obtain multiplication of given expression by using distributive law :

    (a^2 - 9 )(4a) = 4a^3 - 36a

    Question:2 Complete the table


    First expression

    Second expression

    Product

    (i)

    a

    b+c+d

    ...

    (ii)

    x+y-5

    5xy

    ...

    (iii)

    p

    6p^2-7p+5

    ...

    (iv)

    4p^2q^2

    p^2-q^2

    ...

    (v)

    a+b+c

    abc

    ...


    Answer:

    We will use distributive law to find product in each case.


    First expression

    Second expression

    Product

    (i)

    a

    b+c+d

    ab + ac+ ad

    (ii)

    x+y-5

    5xy

    5x^2y + 5xy^2 - 25xy

    (iii)

    p

    6p^2-7p+5

    6p^3 - 7p^2 + 5p

    (iv)

    4p^2q^2

    p^2-q^2

    4p^4q^2 - 4p^2q^4

    (v)

    a+b+c

    abc

    a^2bc + ab^2c + abc^2


    Question:3(i) Find the product.

    (a^2)\times (2a^{22})\times (4a^{26})


    Answer:

    Opening brackets :

    (a^2)\times (2a^{22})\times (4a^{26}) = (a^2\times2a^{22})\times(4a^{26}) = 2a^{24}\times4a^{26}

    or =8a^{50}

    Question:3(ii) Find the product.

    (\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2)

    Answer:

    We have,

    (\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2) = \frac{-3}{5}x^3y^3

    Question:3(iii) Find the product.

    (\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q)

    Answer:

    We have

    (\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q) = -4p^4q^4

    Question:3(iv) Find the product.

    x \times x^2\times x^3\times x^4

    Answer:

    We have x \times x^2\times x^3\times x^4

    x \times x^2\times x^3\times x^4 = (x \times x^2)\times x^3\times x^4

    or (x^3)\times x^3\times x^4

    = x^{10}

    Question:4(a) Simplify 3x(4x-5)+3 and find its values for

    (i) \small x=3

    Answer:

    (a) We have

    3x(4x-5)+3 = 12x^2 - 15x + 3

    Put x = 3,

    We get : 12(3)^2 - 15(3) + 3 = 12(9) - 45 + 3 = 108 - 42 = 66


    Question:4(a) Simplify \small 3x(4x-5)+3 and find its values for

    (ii) \small x=\frac{1}{2}

    Answer:

    We have

    \small 3x(4x-5)+3 = 12x^2 -15x + 3

    Put

    x = \frac{1}{2}

    . So We get,

    12x^2 -15x + 3 = 12(\frac{1}{2})^2 - 15(\frac{1}{2}) + 3 = 6 - \frac{15}{2} = \frac{-3}{2}

    Question:4(b) Simplify \small a(a^2+a+1) + 5 and find its value for

    (i) \small a =0

    Answer:

    We have : \small a(a^2+a+1) +5 = a^3 + a^2 + a +5

    Put a = 0 : = 0^3 + 0^2 + 0 + 5 = 5

    Question:4(b) Simplify \small a(a^2+a+1)+5 and find its value for

    (ii) \small a=1

    Answer:

    We have \small a(a^2+a+1)+5 = a^3 + a^2 + a + 5

    Put a = 1 ,

    we get : 1^3 + 1^2 + 1 + 5 = 1 + 1 + 1+ 5 = 8

    Question:4(b) Simplify \small a(a^2+a+1)+5 and find its value for

    (iii) \small a=-1

    Answer:

    We have \small a(a^2+a+1)+5 .

    or \small a(a^2+a+1)+5 = a^3+a^2+a+5

    Put a = (-1)

    = (-1)^3+(-1)^2+(-1)+5 = -1 + 1 -1 +5 = 4

    Question:5(a) Add: p(p-q),q(q-r) and r(r-p)

    Answer:

    (a)First we will solve each brackets individually.

    p(p-q) = p^2 - pq ; q(q-r) = q^2 - qr ; r(r-p) = r^2 - rp

    Addind all we get : p^2 - pq + q^2 - qr + r^2 - rp

    = p^2 + q^2 + r^2 -pq-qr-rp

    Question:5(b) Add: \small 2x(z-x-y) and \small 2y(z-y-x)

    Answer:

    Firstly, open the brackets:

    \small 2x(z-x-y) = 2xz -2x^2-2xy

    and \small 2y(z-y-x) = 2yz-2y^2-2xy

    Adding both, we get :

    \small 2xz -2x^2-2xy +2yz-2y^2-2xy

    or \small = -2x^2-2y^2-4xy + 2xz+2yz

    Question:5(c) Subtract: \small 3l(l-4m+5n) from \small 4l(10n-3m+2l)

    Answer:

    At first we will solve each bracket individually,

    \small 3l(l-4m+5n) = 3l^2 - 12lm + 15ln

    and \small 4l(10n-3m+2l) = 40ln - 12ml + 8l^2

    Subtracting:

    \small 40ln - 12ml + 8l^2 - (3l^2 - 12lm+15ln)

    or \small = 40ln - 12ml + 8l^2 - 3l^2 + 12lm-15ln

    or \small = 25ln + 5l^2

    Question:5(d) Subtract: \small 3a(a+b+c)-2b(a-b+c) from \small 4c(-a+b+c)

    Answer:

    Solving brackets :

    3a(a+b+c)-2b(a-b+c) = 3a^2+3ab+3ac - 2ab+2b^2-2bc

    = 3a^2+ab+3ac+ 2b^2-2bc

    and \small 4c(-a+b+c) = -4ac +4bc + 4c^2

    Subtracting : \small -4ac +4bc + 4c^2 -(3a^2 + ab + 3ac+2b^2-2bc)

    \small = -4ac + 4bc+4c^2-3a^2-ab-3ac-2b^2+2bc

    \small =-3a^2 -2b^2+4c^2-ab+ 6bc-7ac

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.4

    Question:1(i) Multiply the binomials.

    \small (2x+5) and \small (4x-3)

    Answer:

    We have (2x + 5) and (4x - 3)
    (2x + 5) X (4x - 3) = (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
    = 8 x^{2} - 6x + 20x - 15
    = 8 x^{2} + 14x -15

    Question:1(ii) Multiply the binomials.

    \small (y-8) and \small (3y-4)

    Answer:

    We need to multiply (y - 8) and (3y - 4)
    (y - 8) X (3y - 4) = (y)(3y) + (y)(-4) + (-8)(3y) + (-8)(-4)
    = 3 y^{2} - 4y - 24y + 32
    = 3 y^{2} - 28y + 32

    Question:1(iii) Multiply the binomials

    \small (2.5l-0.5m) and \small (2.5l+0.5m)

    Answer:

    We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
    (2.5l - 0.5m) X (2.5l + 0..5m) = (2.5l)^{2} - (0.5m)^{2} using (a-b)(a+b) = (a)^{2} - (b)^{2}
    = 6.25 l^{2} - 0.25 m^{2}

    Question:1(iv) Multiply the binomials.

    \small (a+3b) and \small (x+5)

    Answer:

    (a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
    = ax + 5a + 3bx + 15b

    Question:1(v) Multiply the binomials.

    \small (2pq+3q^2) and \small (3pq-2q^2)

    Answer:

    (2pq + 3q 2 ) X (3pq - 2q 2 ) = (2pq)(3pq) + (2pq)(-2q 2 ) + ( 3q 2 )(3pq) + (3q 2 )(-2q 2 )
    = 6p 2 q 2 - 4pq 3 + 9pq 3 - 6q 4
    = 6p 2 q 2 +5pq 3 - 6q 4

    Question:1(vi) Multiply the binomials.

    \small (\frac{3}{4}a^2+3b^2) and \small 4(a^2-\frac{2}{3}b^2)

    Answer:

    Multiplication can be done as follows

    \small (\frac{3}{4}a^2+3b^2) X \small (4a^2-\frac{8}{3}b^2) = \frac{3a^{2}}{4} \times 4a^{2} + \frac{3a^{2}}{4} \times (-\frac{8b^{2}}{3}) + 3b^{2} \times 4a^{2} + 3b^{2} \times (-\frac{8b^{2}}{3})


    = 3a^{4} - 2a^{2}b^{2} + 12a^{2}b^{2} - 8b^{4}

    = 3a^{4} + 10a^{2}b^{2} - 8b^{4}

    Question:2(i) Find the product.

    \small (5-2x) \small (3+x)

    Answer:

    (5 - 2x) X (3 + x) = (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
    = 15 + 5x - 6x - 2 x^{2}
    = 15 - x - 2 x^{2}

    Question:2(ii) Find the product.

    \small (x+7y)(7x-y)

    Answer:

    (x + 7y) X (7x - y) = (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
    = 7 x^{2} - xy + 49xy - 7 y^{2}
    = 7 x^{2} + 48xy - 7 y^{2}

    Question:2(iii) Find the product.

    \small (a^2+b)(a+b^2)

    Answer:

    ( a^{2} + b) X (a + b^{2} ) = ( a^{2} )(a) + ( a^{2} )( b^{2} ) + (b)(a) + (b)( b^{2} )
    = a^{3 } + a^{2}b^{2} + ab + b^{3}

    Question:2(iv) Find the product.

    \small (p^2-q^2)(2p+q)

    Answer:

    following is the solution

    ( p^{2}- q^{2} ) X (2p + q) = (p^{2})(2p) + (p^{2})(q) + (-q^{2})(2p) + (-q^{2})(q)
    2p^{3} + p^{2}q - 2q^{2}p - q^{3}

    Question:3(i) Simplify.

    \small (x^2-5)(x+5)+25

    Answer:

    this can be simplified as follows

    ( x^{2} -5) X (x + 5) + 25 = ( x^{2} )(x) + ( x^{2} )(5) + (-5)(x) + (-5)(5) + 25
    = x^{3} + 5x^{2} - 5x -25 + 25
    = x^{3} + 5x^{2} - 5x

    Question:3(ii) Simplify .

    (a^2+5)(b^3+3)+5

    Answer:

    This can be simplified as

    ( a^{2} + 5) X ( b^{3} + 3) + 5 = ( a^{2} )( b^{3} ) + ( a^{2} )(3) + (5)( b^{3} ) + (5)(3) + 5
    = a^{2}b^{3} + 3a^{2} + 5b^{3} + 15+5
    = a^{2}b^{3} + 3a^{2} + 5b^{3} + 20

    Question:3(iii) Simplify.

    (t+s^2)(t^2-s)

    Answer:

    simplifications can be

    (t + s^{2} )( t^{2} - s) = (t)( t^{2} ) + (t)(-s) + ( s^{2} )( t^{2} ) + ( s^{2} )(-s)
    = t^{3} - ts + s^{2}t^{2} - s^{3}

    Question:3(iv) Simplify.

    (a+b)(c-d)+(a-b)(c+d)+2 (ac+bd)

    Answer:

    (a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
    = (a)(c) + (a)(-d) + (b)(c) + (b)(-d) + (a)(c) + (a)(d) + (-b)(c) + (-b)(d) + 2(ac + bd )
    = ac - ad + bc - bd + ac +ad -bc - bd + 2(ac + bd )
    = 2(ac - bd ) + 2(ac +bd )
    = 2ac - 2bd + 2ac + 2bd
    = 4ac

    Question:3(v) Simplify.

    (x+y)(2x+y)+(x+2y)(x-y)

    Answer:

    (x + y) X ( 2x + y) + (x + 2y) X (x - y)
    =(x)(2x) + (x)(y) + (y)(2x) + (y)(y) + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
    = 2 x^{2} + xy + 2xy + y^{2} + x^{2} - xy + 2xy - 2 y^{2}
    =3 x^{2} + 4xy - y^{2}

    Question:3(vi) Simplify.

    (x+y)(x^2-xy+y^2)

    Answer:

    simplification is done as follows

    (x + y) X ( x^{2} -xy + y^{2} ) = x X ( x^{2} -xy + y^{2} ) + y ( x^{2} -xy + y^{2} )
    = x^{3} -x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3}
    = x^{3}+ y^{3}

    Question:3(vii) Simplify.

    (1.5x-4y)(1.5x+4y+3)-4.5x+12y

    Answer:

    (1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y = (1.5x) X (1.5x + 4y + 3) -4y X (1.5x + 4y + 3) - 4.5x + 12y
    = 2.25 x^{2} + 6xy + 4.5x - 6xy - 16 y^{2} - 12y -4.5x + 12 y
    = 2.25 x^{2} - 16 y^{2}

    Question:3(viii) Simplify.

    (a+b+c)(a+b-c)

    Answer:

    (a + b + c) X (a + b - c) = a X (a + b - c) + b X (a + b - c) + c X (a + b - c)
    = a^{2} + ab - ac + ab + b^{2} -bc + ac + bc - c^{2}
    = a^{2} + b^{2} - c^{2} + 2ab

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

    Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

    Answer:

    Identity 1 \Rightarrow (a+b)^{2} = a^{2} + 2ab + b^{2}
    If we replace b with -b in identity 1
    We get,
    a^{2} + 2a(-b) + (-b)^{2} = a^{2} - 2ab + b^{2}
    which is equal to
    (a-b)^{2} which is identity 2
    So, we get identity 2 by replacing b with -b in identity 1

    NCERT Free Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

    Question:1 Verify Identity (IV), for a=2,b=3,x=5 .

    Answer:

    Identity IV
    (a + x)(b + x) = x^{2} + (a+b)x + ab
    So, it is given that a = 2, b = 3 and x = 5
    Lets put these value in identity IV
    (2 + 5)(3 + 5) = 5^{2} + (2 + 3)5 +2 X 3
    7 X 8 = 25 + 5 X 5 + 6
    56 = 25 + 25 + 6
    = 56
    L.H.S. = R.H.S.
    So, by this we can say that identity IV satisfy with given value of a,b and x

    Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity

    Answer:

    Identity IV is \Rightarrow (a +x)(b+x) = x^{2} + (a+b)x + ab
    If a =b than

    (a + x)(a + x) = x^{2} + (a+a)x + a\times a
    (a+x)^{2} = x^{2} + 2ax + a^{2}
    Which is identity I

    Question:3 Consider, the special case of Identity (IV) with a=-c and b=-c What do you get? Is it related to Identity ?

    Answer:

    Identity IV is \Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab
    If a = b = -c than,
    (x - c)(x - c) = x^{2} + (-c + (-c))x + (-c) \times (-c)
    (x-c)^{2} = x^{2} + -2cx + c^{2}
    Which is identity II

    Question:4 Consider the special case of Identity (IV) with b=-a . What do you get? Is it related to Identity?

    Answer:

    Identity IV is \Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab
    If b = -a than,

    (x + a)(x - a) = x^{2} + (a +(-a))x + (-a) \times a
    = x^{2} - a^{2}
    Which is identity III

    Class 8 algebraic expressions and identities NCERT solutions - exercise: 9.5

    Question:1(i) Use a suitable identity to get each of the following products.

    (x+3)(x+3)

    Answer:

    (x + 3) X (x +3) = (x +3)^{2}
    So, we use identity I for this which is
    (a+b)^{2} = a^{2} + 2ab + b^{2}
    In this a=x and b = x
    (x+3)^{2} = x^{2} + 2(x)(3)+ 3^{2}
    = x^{2} + 6x+ 9

    Question:1(ii) Use a suitable identity to get each of the following products in bracket.

    (2y+5)(2y+5)

    Answer:

    (2y + 5) X ( 2y + 5) = (2y +5)^{2}
    We use identity I for this which is
    (a+b)^{2} = a^{2} + 2ab + b^{2}
    IN this a = 2y and b = 5
    (2y+5)^{2} = (2y)^{2} + 2(2y)(5) + 5^{2}
    = (2y+5)^{2} = 4y^{2} + 20y + 25

    Question:1(iii) Use a suitable identity to get each of the following products in bracket.

    (2a-7)(2a-7)

    Answer:

    (2a -7) X (2a - 7) = (2a - 7)^{2}
    We use identity II for this which is
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    in this a = 2a and b = 7
    (2a-7)^{2} = (2a)^{2} - 2(2a)(7) + 7^{2}
    = 4a^{2} - 28a + 49

    Question:1(iv) Use a suitable identity to get each of the following products in bracket.

    (3a - \frac{1}{2}) (3a -\frac{1}{2} )

    Answer:

    (3a - \frac{1}{2}) \times (3a -\frac{1}{2} ) = ((3a - \frac{1}{2}))^{2}
    We use identity II for this which is
    (a-b)^{2} = a^{2} -2ab + b^{2}
    in this a = 3a and b = -1/2
    (3a-\frac{1}{2})^{2} = (3a)^{2} -2(3a)(\frac{1}{2}) + (\frac{1}{2})^{2}
    = 9a^{2} -3a + \frac{1}{4}

    Question:1(v) Use a suitable identity to get each of the following products in bracket.

    (1.1m - 4)(1.1m+4)

    Answer:

    (1.1m - 4)(1.1m+4)
    We use identity III for this which is
    (a - b)(a + b) = a^{2} - b^{2}
    In this a = 1.1m and b = 4
    (1.1m - 4)(1.1m+4) = (1.1m)^{2} - (4)^{2}
    = 1.21 m^{2} - 16

    Question:1(vi) Use a suitable identity to get each of the following products in bracket.

    (a^2+b^2)(-a^2+b^2)

    Answer:

    take the (-)ve sign common so our question becomes
    - -(a^{2}+b^{2})(a^{2}-b^{2})
    We use identity III for this which is
    (a - b)(a + b) = a^{2} - b^{2}
    In this a = a^{2} and b = b^{2}

    -(a^{2}+b^{2})(a^{2}-b^{2}) = -((a^{2})^{2} -(b^{2})^{2}) = -a^{4} + b^{4}

    Question:1(vii) Use a suitable identity to get each of the following.

    (6x-7) (6x+7)

    Answer:

    (6x -7) X (6x - 7) = (6x-7)^{2}
    We use identity III for this which is
    (a - b)(a + b) = a^{2} - b^{2}
    In this a = 6x and b = 7
    (6x -7) X (6x - 7) = (6x)^{2} - (7)^{2} = 36x^{2} - 49

    Question:1(viii) Use a suitable identity to get each of the following product.

    (-a+c)(-a+c)

    Answer:

    take (-)ve sign common from both the brackets So, our question become
    (a -c) X (a -c) = (a -c)^{2}
    We use identity II for this which is
    (a-b)^{2} =a^{2} -2ab + b^{2}
    In this a = a and b = c
    (a-c)^{2} =a^{2} -2ac + c^{2}

    Question:1(ix) Use a suitable identity to get each of the following product.

    (\frac{x}{2}+ \frac{3y}{4})(\frac{x}{2}+ \frac{3y}{4})

    Answer:

    (\frac{x}{2}+ \frac{3y}{4}) \times (\frac{x}{2}+ \frac{3y}{4}) = (\frac{x}{2}+ \frac{3y}{4})^{2}

    We use identity I for this which is
    (a+b)^{2} =a^{2}+2ab + b^{2}
    In this a = \frac{x}{2} and b = \frac{3y}{4}

    (\frac{x}{2}+ \frac{3y}{4})^{2} = (\frac{x}{2})^{2} + 2 (\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^{2}
    = \frac{x^{2}}{4} + \frac{3xy}{4} + \frac{9y^{2}}{16}

    Question:1(x) Use a suitable identity to get each of the following products.

    (7a-9b)(7a-9b)

    Answer:

    (7a-9b) \times (7a-9b) = (7a-9b)^{2}


    We use identity II for this which is
    (a-b)^{2} =a^{2}-2ab + b^{2}
    In this a = 7a and b = 9b
    (7a-9b)^{2} =(7a)^{2}-2(7a)(9b) + (9b)^{2}
    = 49a^{2}-126ab + 81b^{2}

    Question:2(i) Use the identity (x+a) (x+b) = x^2+(a+b)x+ab to find the following products.

    (x+3)(x+7)

    Answer:

    We use identity (x+a) (x+b) = x^2+(a+b)x+ab
    in this a = 3 and b = 7
    (x+3)(x+7) = x^2+(3+7)x+3 \times 7
    = x^2+10x+ 21

    Question:2(ii) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

    (4x+5)(4x+1)

    Answer:

    We use identity (x+a)(x+b)=x^2+(a+b)x+ab
    In this a= 5 , b = 1 and x = 4x
    (4x+5)(4x+1) = (4x)^2+(5+1)4x+(5)(1)
    = 16x^2+24x+5

    Question:2(iii) Use the identity (x+a)(x+b)= x^2+(a+b)x+ab to find the following products.

    (4x-5)(4x-1)

    Answer:

    We use identity (x+a)(x+b)= x^2+(a+b)x+ab
    in this x = 4x , a = -5 and b = -1
    (4x-5)(4x-1) = (4x)^2+(-5-1)4x+(-5)(-1)
    = 16x^2 - 24x+ 5

    Question:1(iv) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

    (4x+5)(4x-1)

    Answer:

    We use identity (x+a)(x+b)=x^2+(a+b)x+ab
    In this a = 5 , b = -1 and x = 4x
    (4x+5)(4x-1) = (4x)^2+(5+(-1))4x+(5)(-1)
    = 16x^2+16x- 5

    Question:2(v) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

    (2x+5y)(2x+3y)

    Answer:

    We use identity (x+a)(x+b)=x^2+(a+b)x+ab
    In this a = 5y , b = 3y and x = 2x
    (2x+5y)(2x+3y) = (2x)^2+(5y+3y)(2x)+(5y)(3y)
    = 4x^2+16xy + 15y^{2}

    Question:2(vi) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

    (2a^2+9)(2a^2+5)

    Answer:

    We use identity (x+a)(x+b)=x^2+(a+b)x+ab
    In this a = 9 , b = 5 and x = 2a^{2}
    (2a^{2}+9)(2a^{2}+5) = (2a^{2})^2+(9+5)2a^{2}+(9)(5)
    = 4a^{4} + 28a^{2} + 45

    Question:2(vii) Use the identity (x+a) (x+b)=x^2+(a+b)x+ab to find the following products.

    (xyz-4) (xyz-2)

    Answer:

    We use identity (x+a)(x+b)=x^2+(a+b)x+ab
    In this a = -4 , b = -2 and x = xyz
    (xyz-4)(xyz-2) = (xyz)^2+((-4)+(-2))xyz+(-4)(-2)
    = x^{2}y^{2}z^{2} -6xyz + 8

    Question:3(i) Find the following squares by using the identities.

    (b-7)^2

    Answer:

    We use identity
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    In this a =b and b = 7
    (b-7)^{2} = b^{2} - 2(b)(7) + 7^{2}
    = b^{2} - 14b + 49

    Question:3(ii) Find the following squares by using the identities.

    (xy+3z)^2

    Answer:

    We use
    (a+b)^{2} = a^{2} + 2ab + b^{2}

    In this a = xy and b = 3z
    (xy+3z)^{2} = (xy)^{2} + 2(xy)(3z) + (3z)^{2}
    = x^{2}y^{2} + 6xyz+ 9z^{2}

    Question:3(iii) Find the following squares by using the identities.

    (6x^2-5y)^2

    Answer:

    We use
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    In this a = 6x^{2} and b = 5y^{2}
    (6x-5y)^{2} = (6x)^{2} - 2(6x)(5y) + (5y)^{2}
    = 36x^{2} - 60xy + 25y^{2}

    Question:3(iv) Find the following squares by using the identities.

    (\frac{2}{3}m+\frac{3}{2}n)^2

    Answer:

    we use the identity
    (a+b)^{2} = a^{2} + 2ab + b^{2}
    In this a = \frac{2m}{3} and b = \frac{3n}{2}
    (\frac{2m}{3} + \frac{3n}{2})^{2} = (\frac{2m}{3})^{2} + 2(\frac{2m}{3})( \frac{3n}{2}) + ( \frac{3n}{2})^{2}

    = \frac{4m^{2}}{9} + 2mn + \frac{9n^{2}}{4}

    Question:3(v) Find the following squares by using the identities.

    (0.4p-0.5q)^2

    Answer:

    we use
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    In this a = 0.4p and b =0.5q
    (0.4p-0.5q)^{2} = (0.4p)^{2} - 2(0.4p)(0.5q) + (0.5q)^{2}
    = 0.16p^{2} - 0.4pq + 0.25q^{2}

    Question:3(vi) Find the following squares by using the identities.

    (2xy+5y)^2

    Answer:

    we use the identity
    (a+b)^{2} = a^{2} + 2ab + b^{2}
    In this a = 2xy and b =5y
    (2xy+5y)^{2} = (2xy)^{2} + 2(2xy)(5y) + (5y)^{2}
    = 4x^{2}y^{2} + 20xy^{2} + 25y^{2}

    Question:4(i) Simplify:

    (a^2-b^2)^2

    Answer:

    we use
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    In this a = a^{2} and b = b^{2}
    (a^{2}-b^{2})^{2} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}
    = a^{4} - 2a^{2}b^{2} + b^{4}

    Question:4(ii) Simplify.

    (2x+5)^2-(2x-5)^2

    Answer:

    we use
    a^{2} - b^{2} = (a-b)(a+b)
    In this a = (2x + 5) and b = (2x - 5)
    (2x + 5)^{2} - (2x - 5)^{2} = ( (2x + 5)- (2x - 5))( (2x + 5)+ (2x - 5))
    = ( 2x + 5- 2x + 5)( 2x + 5+ 2x - 5)
    = (4x)(10)
    =40x

    or

    remember that

    (a+b)^2-(a-b)^2=4ab

    here a= 2x, b= 5

    4ab=4\times 2x \times 5=40x

    Question:4(iii) Simplify.

    (7m-8n)^2+(7m+8n)^2

    Answer:

    we use
    (a-b)^{2} = a^{2} -2ab + b^{2} and (a+b)^{2} = a^{2} +2ab + b^{2}
    In this a = 7m and b = 8n
    (7m-8n)^{2} = (7m)^{2} -2(7m)(8n) + (8n)^{2}
    = 49m^{2} -112mn + 64n^{2}
    and
    (7m+8n)^{2} = (7m)^{2} +2(7m)(8n) + (8n)^{2}
    = 49m^{2} +112mn + 64n^{2}

    So, (7m - 8n)^{2} + (7m + 8n)^{2} = 49m^{2} -112mn + 64n^{2} + 49m^{2} +112mn + 64n^{2}
    = 2(49m^{2} + 64n^{2})

    or

    remember that

    (a-b)^2+(a+b)^2=2(a^2+b^2)

    Question: 4(iv) Simplify.

    (4m+5n)^2+(5m+4n)^2

    Answer:

    we use
    (a+b)^{2} = a^{2} +2ab + b^{2}
    1 ) In this a = 4m and b = 5n

    (4m+5n)^{2} = (4m)^{2} +2(4m)(5n) + (5n)^{2}
    = 16m^{2} +40mn + 25n^{2}
    2 ) in this a = 5m and b = 4n
    (5m+4n)^{2} = (5m)^{2} +2(5m)(4n) + (4n)^{2}
    = 25m^{2} +40mn + 16n^{2}

    So, (4m + 5n)^{2} + (5m + 4n)^{2} = 16m^{2} +40mn + 25n^{2} + 25m^{2} +40mn + 16n^{2}
    = 41m^{2} +80mn + 41n^{2}

    Question: 4(v) Simplify.

    (2.5p-1.5q)^2-(1.5p-2.5q)^2

    Answer:

    we use
    a^{2}- b^{2} = (a-b)(a+b)
    1 ) In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)
    (2.5p- 1.5q)^{2}- (1.5p- 2.5q)^{2} = ( (2.5p- 1.5q)- (1.5p- 2.5q))( (2.5p- 1.5q)+ (1.5p- 2.5q))
    = ( 2.5p- 1.5q- 1.5p + 2.5q)(2.5p- 1.5q+ 1.5p- 2.5q)
    = 4(p + q ) (p - q)
    = 4 (p^{2} - q^{2})

    Question:4(vi) Simplify.

    (ab+bc)^2-2ab^2c

    Answer:

    We use identity
    (a+b)^{2} = a^{2} + 2ab + b^{2}
    In this a = ab and b = bc
    (ab+bc)^{2} = (ab)^{2} + 2(ab)(bc) + (bc)^{2}
    = a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}
    Now, a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2} - 2ab^{2}c
    = a^{2}b^{2} + b^{2}c^{2}

    Question:4(vii) Simplify.

    (m^2 -n^2m)^2+2m^3n^2

    Answer:

    We use identity
    (a-b)^{2} = a^{2} - 2ab + b^{2}
    In this a = m^{2} and b = n^{2}m
    (m^{2}-n^{2}m)^{2} = (m^{2})^{2} - 2(m^{2})(n^{2}m) + (n^{2}m)^{2}
    = m^{4} - 2m^{3}n^{2} + n^{4}m^{2}
    Now, m^{4} - 2m^{3}n^{2} + n^{4}m^{2} + 2m^{3}n^{2}
    = m^{4} + n^{4}m^{2}

    Question:5(i) Show that

    (3x+7)^2-84x=(3x-7)^2

    Answer:

    L.H.S. = (3x+7)^2 - 84x = 9x^2 + 42x + 49 - 84x

    = 9x^2 - 42 x +49

    = (3x - 7)^2

    = R.H.S.

    Hence it is prooved

    Question:5(ii) Show that

    (9p-5q)^2+180pq=(9p+5q)^2

    Answer:

    L.H.S. = (9p-5q)^2+180pq = 81p^2 - 90pq + 25q^2 + 180pq (Using (a-b)^2 = a^2 - 2ab + b^2 )

    = 81p^2 +90pq + 25q^2

    = (9p + 5q)^2 \left ( (a+b)^2 = a^2 + 2ab + b^2 \right )

    = R.H.S.

    Question:5(iii) Show that.

    (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn=\frac{16}{9}m^2+\frac{9}{16}n^2

    Answer:

    First we will solve the LHS :

    = (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn

    or = \frac{16}{9}m^2 + \frac{9}{16}n^2

    = RHS

    Question:5(iv) Show that.

    (4pq+3q)^2-(4pq-3q)^2=48pq^2

    Answer:

    Opening both brackets we get,

    (4pq+3q)^2-(4pq-3q)^2 = 16p^2q^2 + 24pq^2 + 9q^2 - (16p^2q^2 - 24pq^2 + 9q^2)

    = 16p^2q^2 + 24pq^2 + 9q^2 - 16p^2q^2 + 24pq^2 - 9q^2)

    = 48pq^2

    = R.H.S.

    Question:5(v) Show that

    (a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)=0

    Answer:

    Opening all brackets from the LHS, we get :

    (a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)\\\\ =\ a^2 +ab - ab- b^2 + b^2+bc - bc -c^2 + c^2 +ca - ac -a^2

    = 0 = RHS

    Question:6(i) Using identities, evaluate.

    71^2

    Answer:

    We will use the identity:

    (a + b)^2 = a^2 + 2ab + b^2

    So, 71^2 = (70 +1)^2 = 70^2 + 2(70)(1) + 1^2

    = 4900 + 140 + 1

    = 5041

    Question:6(ii) Using identities, evaluate.

    99^2

    Answer:

    Here we will use the identity :

    (a - b)^2 = a^2 - 2ab + b^2

    So : 99^2 = (100 - 1) ^2 = 100^2 - 2(100)(1) + 1^2

    or = 10000 - 200 + 1

    = 9801

    Question:6(iii) Using identities, evaluate.

    102^2

    Answer:

    Here we will use the identity :

    (a+b)^2 = a^2 +2ab + b^2

    So :

    102^2 = (100 + 2)^2 = 100^2 + 2(100)(2) + 2^2

    or = 10000 + 400 + 4

    = 10404

    Question:6(iv) Using identities, evaluate.

    998^2

    Answer:

    Here we will the identity :

    998^2=(1000 - 2)^2 = 1000^2 - 2(1000)(2) + 2^2

    or = 1000000 - 4000+ 4

    or = 996004

    Question:6(v) Using identities, evaluate.

    5.2^2

    Answer:

    Here we will use :

    (a + b)^2 = a^2 + 2ab + b^2

    Thus

    (5.2)^2 = (5.0 + 0.2)^2 = 5^2 + 2(5)(0.2) + (0.2)^2

    or = 25 + 2 + 0.04

    = 27.04

    Question:6(vi) Using identities, evaluate.

    297 \times 303

    Answer:

    This can be written as :

    297\times303 = (300-3)\times(300+3)

    using (a-b)(a+b)=a^2-b^2

    or = 90000 - 9

    = 89991

    Question:6(vii) Using identities, evaluate.

    78 \times 82

    Answer:

    This can be written in form of :

    78\times82 = (80 - 2) \times(80+2)

    or = 80^2 - 2^2 \because \left ( a-b \right )\left ( a+b \right ) = a^2 - b^2

    or = 6400- 4 = 6396

    Question:6(viii) Using identities, evaluate.

    8.9^2

    Answer:

    Here we will use the identity :

    (a - b)^2 = a^2 - 2ab + b^2

    Thus :

    8.9^2 = (9 - 0.1) ^2 = 9^2 - 2(9)(0.1) + 0.1^2

    or = 81 - 1.8 + 0.01

    or = 79.21

    Question:6(ix) Using identities, evaluate.

    10.5\times9.5

    Answer:

    This can be written as :

    10.5\times9.5 = (10 +0.5)\times(10-0.5)

    or = 10^2 - 0.5^2 \because (a+b)(a-b) = a^2 - b^2

    or = 100 - 0.25

    or = 99.75

    Question:7(i) Using a^2-b^2=(a+b)(a-b) , find

    51^2-49^2

    Answer:

    We know,

    a^2-b^2=(a+b)(a-b)

    Using this formula,

    51^2-49^2 = (51 + 49)(51 - 49)

    = (100)(2)

    = 200

    Question:7(ii) Using a^2-b^2=(a+b)(a-b) , find

    (1.02)^2-(0.98)^2

    Answer:

    We know,

    a^2-b^2=(a+b)(a-b)

    Using this formula,

    (1.02)^2-(0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)
    = (2.00)(0.04)

    = 0.08

    Question:7(iii) Using a^2-b^2=(a+b)(a-b) , find.

    153^2-147^2

    Answer:

    We know,

    a^2-b^2=(a+b)(a-b)

    Using this formula,

    153^2-147^2 = (153 - 147)(153 +147)

    =(6) (300)

    = 1800

    Question:7(iv) Using a^2-b^2=(a+b )(a-b) , find

    12.1^2-7.9^2

    Answer:

    We know,

    a^2-b^2=(a+b)(a-b)

    Using this formula,

    (1.02)^2-(0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)

    = (2.00)(0.04)

    = 0.08

    Question:8(i) Using (x+a)(x+b)=x^2+(a+b)x+ab 103 \times 104

    Answer:

    We know,

    (x+a)(x+b)=x^2+(a+b)x+ab

    Using this formula,

    103 \times 104 = (100 + 3)(100 + 4)

    Here x =100, a = 3, b = 4

    \therefore 103 \times 104 = 100^2+(3+4)100+(3\times4)

    = 10000+1200+12

    = 11212

    Question:8(ii) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

    5.1\times 5.2

    Answer:

    We know,

    (x+a)(x+b)=x^2+(a+b)x+ab

    Using this formula,

    5.1\times 5.2 = (5 + 0.1)(5 + 0.2)

    Here x =5, a = 0.1, b = 0.2

    \therefore 5.1\times 5.2 =5^2+(0.1 + 0.2)5+(0.1\times0.2)

    = 25+1.5+0.02

    = 26.52

    Question:8(iii) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

    103 \times 98

    Answer:

    We know,

    (x+a)(x+b)=x^2+(a+b)x+ab

    Using this formula,

    103 \times 98 = (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

    Here x =100, a = 3, b = -2

    \therefore 103 \times 98 = 100^2+(3 + (-2))100+(3\times(-2))

    = 10000+100-6

    = 10094

    Question: 8(iv) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

    9.7 \times 9.8

    Answer:

    We know,

    (x+a)(x+b)=x^2+(a+b)x+ab

    Using this formula,

    9.7 \times 9.8 = (10 - 0.3)(10 - 0.2) = {10 + (-0.3)}{10 + (-0.2)}

    Here x =10, a = -0.3, b = -0.2

    \therefore 9.7 \times 9.8 = 10^2+((-0.3) + (-0.2))10+((-0.3)\times(-0.2))

    = 100-5+0.06

    = 95.

    NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topics

    • What are Expressions?
    • Terms, Factors and Coefficients
    • Monomials, Binomials,f and Polynomials
    • Like and Unlike Terms
    • Addition and Subtraction of Algebraic Expressions
    • Multiplication of Algebraic Expressions: Introduction
    • Multiplying a Monomial by a Monomial
    • Multiplying two monomials
    • Multiplying three or more monomials
    • Multiplying a Monomial by a Polynomial
    • Multiplying a monomial by a binomial
    • Multiplying a monomial by a trinomial
    • Multiplying a Polynomial by a Polynomial
    • Multiplying a binomial by a binomial
    • Multiplying a binomial by a trinomial
    • What is an Identity?
    • Standard Identities
    • Applying Identities

    NCERT Solutions for Class 8 Maths - Chapter Wise

    NCERT Solutions for Class 8 - Subject Wise

    Some Important Identities From NCERT Book for Class 8 Chapter 9 Algebraic Expressions And Identities

    • (a+b)^2=(a^2+2ab+b^2)

    you can write

    (a+b)^2=(a+b)(a+b)

    Can be simplified as follows

    (a+b)(a+b)=a\times a+a\times b+a\times b+b\times b

    Now add each term

    a\times a+a\times b+a\times b+b\times b=a^2+ab+ab+b^2

    =a^2+2ab+b^2

    (a+b)^2=(a^2+2ab+b^2)

    • (a-b)^2=(a^2-2ab+b^2)
    • (a-b)(a+b)=a^2-b^2
    • (x+a)(x+b)=x^{2}+(a+b) x+a b

    You can form the above identities by yourself. These above identities have been used in many problems of NCERT solutions for class 8 maths chapter 9 algebraic expression and identities.

    Also Check NCERT Books and NCERT Syllabus here

    Frequently Asked Question (FAQs)

    1. What are the important topics of NCERT syllabus chapter Algebraic Expressions and Identities ?

    Addition and subtraction of the algebraic expression, multiplication of the algebraic expression, standard identities, and application of identities are important topics in this chapter.

    2. Does CBSE class maths is tough ?

    CBSE class 8 maths is not tough at all. It teaches a very basic and simple maths.

    3. How many chapters are there in the CBSE class 8 maths ?

    There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

    4. Where can I find the complete solutions of NCERT for class 8 ?

    Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

    5. Where can I find the complete solutions of NCERT for class 8 maths ?

    Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

    6. Which is the official website of NCERT ?

    NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

    Articles

    Get answers from students and experts

    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    Ophthalmologist

    Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment. 

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    Multimedia Specialist

    A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

    2 Jobs Available
    Visual Communication Designer

    Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Advertising Manager

    Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

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    3 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Quality Systems Manager

    A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party. 

    2 Jobs Available
    Merchandiser

    A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying. 

    2 Jobs Available
    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

    2 Jobs Available
    Production Planner

    Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner. 

    2 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    DevOps Architect

    A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties. 

    2 Jobs Available
    Cloud Solution Architect

    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

    2 Jobs Available
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