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

Can you calculate (102 × 102) without using any pen or paper, or calculate $(a+b{)}^2$ in a moment? To do this, you don't need any magic tricks, but only have to learn the concept of Algebraic Expressions and Identities. In the 8th chapter of the NCERT Class 8 Maths, you will find Algebraic Expressions and Identities. In this chapter, you will learn how to build or break down algebraic expressions, understand the logic behind the identities, and take your algebraic skills to new heights.

This Story also Contains

1. Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download
2. Algebraic Expressions and Identities Class 8 Solutions - Important Formulae
3. Algebraic Expressions and Identities Class 8 NCERT Solutions
4. NCERT Solutions For Class 8 Maths - Chapter Wise
5. Importance of Solving NCERT Questions of Class 8 Maths Chapter 8
6. NCERT Class 8 Maths Solutions: Subject Wise
7. NCERT Books and Syllabus for Class 8

This article on NCERT solutions for class 8 Maths Chapter 8 Algebraic Expressions and Identities offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 8 Maths. Students who are in need of Algebraic Expressions and Identities class 8 solutions will find this article very useful. It covers all the important Class 8 Maths Chapter 8 question answers. These Algebraic Expressions and Identities class 8 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 8 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

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

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)

Algebraic Expressions and Identities Class 8 NCERT Solutions

NCERT Class 8 Solution Exercise: 8.1
Page number: 94, Total questions: 2

Question: 1(i) Add the following.

$ab-bc,bc-ca,ca-ab$

Answer:

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

Question: 1(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: 1(iii) Add the following

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

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: 1(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: 2(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: 2(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: 2(c) Subtract $4p^{2}q-3pq+5p{q}^2-8p+7q-10$ from $18-3p-11q+5pq-2p{q}^2+5p^{2}q$

Answer:

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

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

$=28+5p-18q+8pq-7p{q}^2+p^{2}q$

Class 8 maths chapter 8 question answer - exercise: 8.2
Page number: 97-98, Total questions: 5

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

$4,7p$

Answer:

$4\times 7p=28p$

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

$-4p,7p$

Answer:

$$\begin{gathered} -4p\times 7p \\ =(-4\times 7)p\times p \\ =-28p^2 \end{gathered}$$

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

$-4p,7pq$

Answer:

$$\begin{gathered} -4p\times 7pq \\ =(-4\times 7)p\times pq \\ =-28p^{2}q \end{gathered}$$

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

$4p^3,-3p$

Answer:

$$\begin{gathered} 4p^3\times (-3p) \\ =4\times (-3)p^3\times p \\ =-12p^4 \end{gathered}$$

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

$4p,0$

Answer:

$4p\times 0=0$

Question: 2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(p,q)$

Answer:

The question can be solved as follows

$$\begin{gathered} Area=length\times breadth \\ =(p\times q) \\ =pq \end{gathered}$$

Question: 2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth, respectively.

$(10m,5n)$

Answer:

The area is calculated as follows

$$\begin{gathered} Area=length\times breadth \\ =10m\times 5n \\ =50mn \end{gathered}$$

Question: 2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(20x^2,5y^2)$

Answer:

The following is the solution

$$\begin{gathered} Area=length\times breadth \\ =20x^2\times 5y^2 \\ =100x^2{y}^2 \end{gathered}$$

Question: 2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(4x,3x^2)$

Answer:

The area of a rectangle is

$$\begin{gathered} Area=length\times breadth \\ =4x\times 3x^2 \\ =12x^3 \end{gathered}$$

Question: 2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths, respectively.

$(3mn,4np)$

Answer:

The area is calculated as follows

$$\begin{gathered} Area=length\times breadth \\ =3mn\times 4np \\ =12m{n}^{2}p \end{gathered}$$.

Question: 3. Complete the table of products.

Answer:

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

$5a,3a^2,7a^4$

Answer:

$$\begin{gathered} Volume=length\times breadth\times height \\ =5a\times 3a^2\times 7a^4 \\ =15a^3\times 7a^4 \\ =105a^7 \end{gathered}$$

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

$$\begin{gathered} Volume=length\times breadth\times height \\ =2p\times 4q\times 8r \\ =8pq\times 8r \\ =64pqr \end{gathered}$$

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

$xy,2x^{2}y,2x{y}^2$

Answer:

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

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

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

$$\begin{gathered} Volume=length\times breadth\times height \\ =a\times 2b\times 3c \\ =2ab\times 3c \\ =6abc \end{gathered}$$

Question: 5(i) Obtain the product of

$xy,yz,zx$

Answer:

The product

$$\begin{gathered} xy\times yz\times zx \\ =x{y}^{2}z\times zx \\ =x^2{y}^2{z}^2 \end{gathered}$$

Question: 5(ii) Obtain the product of

$a,-a^2,a^3$

Answer:

The product

$$\begin{gathered} a\times (-a^2)\times a^3 \\ =-a^{{}^3}\times a^3=-a^6 \end{gathered}$$

Question: 5(iii) Obtain the product of

$2,4y,8y^2,16y^3$

Answer:

The product

$$\begin{gathered} 2\times 4y\times 8y^2\times 16y^3 \\ =8y\times 8y^2\times 16y^3 \\ =64y^3\times 16y^3 \\ =1024y^6 \end{gathered}$$

Question: 5(iv) Obtain the product of

$a,2b,3c,6abc$

Answer:

The product

$$\begin{gathered} a\times 2b\times 3c\times 6abc \\ =2ab\times 3c\times 6abc \\ =6abc\times 6abc \\ =36a^2{b}^2{c}^2 \end{gathered}$$

Question: 5(v) Obtain the product of

$m,-mn,mnp$

Answer:

The product

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

Class 8 maths chapter 8 NCERT solutions - Topic 8.8.1: Multiplying a Monomial by a Binomial

Question: (i) Find the product

$2x(3x+5xy)$

Answer:

Using distributive law,

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

Question: (ii) Find the product

$a^2(2ab-5c)$

Answer:

Using distributive law,

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

NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities - Topic 8.8.2 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 8 question answer - exercise: 8.3
Page number: 100, Total questions: 5

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,

$ab(a-b)=a^{2}b-a{b}^2$

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

$a+b,7a^2{b}^2$

Answer:

Using distributive law, we can obtain the multiplication of the given expression:

$(a+b)(7a^2{b}^2)=7a^3{b}^2+7a^2{b}^3$

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

$a^2-9,4a$

Answer:

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

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

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

$pq+qr+rp,0$

Answer:

Using distributive law :

$(pq+qr+rp)(0)=pq(0)+qr(0)+rp(0)=0$

Question: 2 Complete the table

Answer:

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

Question: 3(i) Find the product.

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

Answer:

Opening brackets :

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

or $=8a^{50}$

Question: 3(ii) Find the product.

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

Answer:

We have,

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

Question: 3(iii) Find the product.

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

Answer:

We have

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

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) $x=3$

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 $3x(4x-5)+3$ and find its values for

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

Answer:

We have

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

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 $a(a^2+a+1)+5$ and find its value for

(i) $a=0$

Answer:

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

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

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

(ii) $a=1$

Answer:

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

Put a = 1 ,

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

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

(iii) $a=-1$

Answer:

We have $a(a^2+a+1)+5$ .

or $a(a^2+a+1)+5=a^3+a^2+a+5$

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 bracket 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: $2x(z-x-y)$ and $2y(z-y-x)$

Answer:

Firstly, open the brackets:

$2x(z-x-y)=2xz-2x^2-2xy$

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

Adding both, we get :

$2xz-2x^2-2xy+2yz-2y^2-2xy$

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

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

Answer:

At first we will solve each bracket individually,

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

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

Subtracting:

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

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

or $=25ln+5l^2$

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

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 $4c(-a+b+c)=-4ac+4bc+4c^2$

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

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

$=-3a^2-2b^2+4c^2-ab+6bc-7ac$.

Class 8 maths chapter 8 question answer - exercise: 8.4
Page number: 102, Total questions: 3

Question: 1(i) Multiply the binomials.

$(2x+5)$ and $(4x-3)$

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.

$(y-8)$ and $(3y-4)$

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

$(2.5l-0.5m)$ and $(2.5l+0.5m)$

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.

$(a+3b)$ and $(x+5)$

Answer:

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

Question:1(v) Multiply the binomials.

$(2pq+3q^2)$ and $(3pq-2q^2)$

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.

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

Answer:

Multiplication can be done as follows

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

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

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

Question: 2(i) Find the product.

$(5-2x)$ $(3+x)$

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.

$(x+7y)(7x-y)$

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.

$(a^2+b)(a+b^2)$

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.

$(p^2-q^2)(2p+q)$

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.

$(x^2-5)(x+5)+25$

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

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

Importance of Solving NCERT Questions of Class 8 Maths Chapter 8

• Solving these NCERT questions will help students understand the basic concepts of Algebraic Expressions and Identities easily.
• Students can practice various types of questions, which will improve their problem-solving skills.
• These NCERT exercises cover all the essential topics and concepts so that students can be well-prepared for various exams.
• By solving these NCERT problems, students will learn about all the real-life applications of Algebraic Expressions and Identities.

NCERT Class 8 Maths Solutions: Subject Wise

Students can use the links below to prepare efficiently in other subjects as well as Mathematics.

• NCERT solutions for class 8 maths
• NCERT solutions for class 8 science

NCERT Books and Syllabus for Class 8

The following links will give students access to the latest CBSE syllabus and some reference books.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8

Keep Working Hard and Happy Learning!