ALLEN Coaching
ApplyRegister for ALLEN Scholarship Test & get up to 90% Scholarship
An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. These algebraic identities are also used in the factorisation of polynomials. These identities are used to solve different types of algebraic equations and polynomials. Algebraic identities are used to solve not only algebraic equations but also to solve geometry, trigonometry expressions by providing shortcuts. These identities help to simplify the complex equations.
Class 9 Maths chapter 2 exercise 2.4 includes a variety of problems related to the application of all the algebraic identities in the question. In exercise 2.5, Class 9 Maths a lot of problems which also include real-life applications. The 9th class maths exercise 2.4 answers have been meticulously prepared by subject experts and are presented in a comprehensive and easily understandable manner. This exercise consists of a total of sixteen questions, each containing multiple parts. Students can readily access NCERT Solutions class 9 maths chapter 2 exercise 2.4 in PDF format, allowing them to use them offline without requiring an internet connection, and they are made available free of charge. Additionally, along with Exercise 2.4 class 9 maths, you'll also find NCERT Books to study all the chapters and other subjects in a detailed manner.
Q1 (i) Use suitable identities to find the following product: $(x + 4) ( x + 10)$
Answer:
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = 4 \ \ and \ \ b = 10$
$(x+4)(x+10)= x^2+(4 + 10)x+4\times 10$
$= x^2+14x+40$
Therefore, $(x + 4) ( x + 10)$ is equal to $x^2+14x+40$
Q1 (ii) Use suitable identities to find the following product: $(x+8)(x-10)$
Answer:
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = 8 \ \ and \ \ b = -10$
$(x+8)(x-10)= x^2+(8 + (-10))x+(-10)\times 8$
$= x^2-2x-80$
Therefore, $(x+8)(x-10)$ is equal to $x^2-2x-80$
Q1 (iii) Use suitable identities to find the following product: $(3x+4)(3x - 5)$
Answer:
We can write $(3x+4)(3x - 5)$ as
$(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}$
$9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )$
$=9x^2-3x-20$
Therefore, $(3x+4)(3x - 5)$ is equal to $9x^2-3x-20$
Q1 (iv) Use suitable identities to find the following product: $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$
Answer:
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $x=y^2 \ \ and \ \ a = \frac{3}{2}$
$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2$
$= y^4-\frac{9}{4}$
Therefore, $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$ is equal to $y^4-\frac{9}{4}$
Q1 (v) Use suitable identities to find the following product: $(3 - 2x) (3 + 2x)$
Answer:
We can write $(3 - 2x) (3 + 2x)$ as
$(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )$
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $a = \frac{3}{2}$
$-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )$
$=9-4x^2$
Therefore, $(3 - 2x) (3 + 2x)$ is equal to $9-4x^2$
Q2 (i) Evaluate the following product without multiplying directly: $103 \times 107$
Answer:
We can rewrite $103 \times 107$ as
$\Rightarrow 103 \times 107= (100+3)\times (100+7)$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $x =100 , a=3 \ \ and \ \ b = 7$
$(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7$
$=10000+1000+21= 11021$
Therefore, value of $103 \times 107$ is $11021$
Q2 (ii) Evaluate the following product without multiplying directly: $95 \times 96$
Answer:
We can rewrite $95 \times 96$ as
$\Rightarrow 95 \times 96= (100-5)\times (100-4)$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $x =100 , a=-5 \ \ and \ \ b = -4$
$(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)$
$=10000-900+20= 9120$
Therefore, value of $95 \times 96$ is $9120$
Q2 (iii) Evaluate the following product without multiplying directly: $104 \times 96$
Answer:
We can rewrite $104 \times 96$ as
$\Rightarrow 104 \times 96= (100+4)\times (100-4)$
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $x =100 \ \ and \ \ a=4$
$(100+4)\times (100-4)= (100)^2-(4)^2$
$=10000-16= 9984$
Therefore, value of $104 \times 96$ is $9984$
Q3 (i) Factorise the following using appropriate identities: $9x^2 + 6xy + y^2$
Answer:
We can rewrite $9x^2 + 6xy + y^2$ as
$\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2$
Using identity $\Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2$
Here, $a= 3x \ \ and \ \ b = y$
Therefore,
$9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)$
Q3 (ii) Factorise the following using appropriate identities: $4y^2 - 4y + 1$
Answer:
We can rewrite $4y^2 - 4y + 1$ as
$\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2$
Using identity $\Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2$
Here, $a= 2y \ \ and \ \ b = 1$
Therefore,
$4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)$
Q3 (iii) Factorise the following using appropriate identities: $x^2 - \frac{y^2}{100}$
Answer:
We can rewrite $x^2 - \frac{y^2}{100}$ as
$\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2$
Using identity $\Rightarrow a^2-b^2 = (a-b)(a+b)$
Here, $a= x \ \ and \ \ b = \frac{y}{10}$
Therefore,
$x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )$
Q4 (i) Expand each of the following, using suitable identities: $(x + 2y+4z)^2$
Answer:
Given is $(x + 2y+4z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = x , b = 2y \ \ and \ \ c = 4z$
Therefore,
$(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x$
$= x^2+4y^2+16z^2+4xy+16yz+8zx$
Q4 (ii) Expand each of the following, using suitable identities: $(2x - y + z)^2$
Answer:
Given is $(2x - y + z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = 2x , b = -y \ \ and \ \ c = z$
Therefore,
$(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x$
$= 4x^2+y^2+z^2-4xy-2yz+4zx$
Q4 (iii) Expand each of the following, using suitable identities: $(-2x + 3y + 2z)^2$
Answer:
Given is $(-2x + 3y + 2z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a =- 2x , b = 3y \ \ and \ \ c = 2z$
Therefore,
$(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)$
$= 4x^2+9y^2+4z^2-12xy+12yz-8zx$
Q4 (iv) Expand each of the following, using suitable identities: $(3a - 7b - c)^2$
Answer:
Given is $(3a - 7b - c)^2$
We will Use identity
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$
Here, $x =3a , y = -7b \ \ and \ \ z = -c$
Therefore,
$(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c)$ $.3a$
$= 9a^2+49b^2+c^2-42ab+14bc-6ca$
Q4 (v) Expand each of the following, using suitable identities: $(-2x + 5y -3z)^2$
Answer:
Given is $(-2x + 5y -3z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a =- 2x , b = 5y \ \ and \ \ c = -3z$
Therefore,
$(-2x +5y-3z)^2$ $= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)$
$= 4x^2+25y^2+9z^2-20xy-30yz+12zx$
Answer:
Given is $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$
We will Use identity
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$
Here, $x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1$
Therefore,
$\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$ $=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )$
$= \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$
Q5 (i) Factorise: $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$
Answer:
We can rewrite $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ as
$\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ $= (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = 2x , b = 3y \ \ and \ \ c = -4z$
Therefore,
$4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x+3y-4z)^2$
$= (2x+3y-4z)(2x+3y-4z)$
Q5 (ii) Factorise: $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$
Answer:
We can rewrite $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ as
$\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ $= (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z$
Therefore,
$2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2$
$=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)$
Q6 (i) Write the following cubes in expanded form: $(2x + 1)^3$
Answer:
Given is $(2x + 1)^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a= 2x \ \ and \ \ b= 1$
Therefore,
$(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2$
$= 8x^3+1+12x^2+6x$
Q6 (ii) Write the following cube in expanded form: $(2a-3b)^3$
Answer:
Given is $(2a-3b)^3$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x= 2a \ \ and \ \ y= 3b$
Therefore,
$(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2$
$= 8a^3-9b^3-36a^2b+54ab^2$
Q6 (iii) Write the following cube in expanded form: $\left[\frac{3}{2}x + 1\right ]^3$
Answer:
Given is $\left[\frac{3}{2}x + 1\right ]^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a= \frac{3x}{2} \ \ and \ \ b= 1$
Therefore,
$\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2$
$= \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}$
Q6 (iv) Write the following cube in expanded form: $\left[x - \frac{2}{3} y\right ]^3$
Answer:
Given is $\left[x - \frac{2}{3} y\right ]^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=x \ and \ \ b= \frac{2y}{3}$
Therefore,
$\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2$
$= x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}$
Q7 (i) Evaluate the following using suitable identities: $(99)^3$
Answer:
We can rewrite $(99)^3$ as
$\Rightarrow (99)^3=(100-1)^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=100 \ and \ \ b= 1$
Therefore,
$(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2$
$= 1000000-1-30000+300= 970299$
Q7 (ii) Evaluate the following using suitable identities: $(102)^3$
Answer:
We can rewrite $(102)^3$ as
$\Rightarrow (102)^3=(100+2)^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a=100 \ and \ \ b= 2$
Therefore,
$(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2$
$= 1000000+8+60000+1200= 1061208$
Q7 (iii) Evaluate the following using suitable identities: $(998)^3$
Answer:
We can rewrite $(998)^3$ as
$\Rightarrow (998)^3=(1000-2)^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=1000 \ and \ \ b= 2$
Therefore,
$(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2$
$= 1000000000-8-6000000+12000= 994011992$
Q8 (i) Factorise the following: $8a^3 + b^3 + 12a^2 b + 6ab^2$
Answer:
We can rewrite $8a^3 + b^3 + 12a^2 b + 6ab^2$ as
$\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2$ $= (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2$
We will use identity
$(x+y)^3=x^3+y^3+3x^2y+3xy^2$
Here, $x=2a \ \ and \ \ y= b$
Therefore,
$8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3$
$=(2a+b)(2a+b)(2a+b)$
Q8 (ii) Factorise the following: $8a ^3 - b^3 - 12a^2 b + 6ab^2$
Answer:
We can rewrite $8a ^3 - b^3 - 12a^2 b + 6ab^2$ as
$\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2$ $= (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=2a \ \ and \ \ y= b$
Therefore,
$8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3$
$=(2a-b)(2a-b)(2a-b)$
Q8 (iii) Factorise the following: $27 - 125a^ 3 - 135a + 225a^2$
Answer:
We can rewrite $27 - 125a^ 3 - 135a + 225a^2$ as
$= (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=3 \ \ and \ \ y= 5a$
Therefore,
$27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3$
$=(3-5a)(3-5a)(3-5a)$
Q8 (iv) Factorise the following: $64a^3 - 27b^3 - 144a^2 b + 108ab^2$
Answer:
We can rewrite $64a^3 - 27b^3 - 144a^2 b + 108ab^2$ as
$\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2$ $= (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=4a \ \ and \ \ y= 3b$
Therefore,
$64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2$
$=(4a-3b)(4a-3b)(4a-3b)$
Q8 (v) Factorise the following: $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$
Answer:
We can rewrite $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ as
$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=3p \ \ and \ \ y= \frac{1}{6}$
Therefore,
$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$
$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$
Q9 (i) Verify: $x^3 + y^3 = (x +y)(x^2 - xy + y^2)$
Answer:
We know that
$(x+y)^3=x^3+y^3+3xy(x+y)$
Now,
$\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)$
$\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )$
$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )$ $(\because (a+b)^2=a^2+b^2+2ab)$
$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )$
Hence proved.
Q9 (ii) Verify: $x^3 - y^3 = (x -y)(x^2 + xy + y^2)$
Answer:
We know that
$(x-y)^3=x^3-y^3-3xy(x-y)$
Now,
$\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)$
$\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )$ $(\because (a-b)^2=a^2+b^2-2ab)$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )$
Hence proved.
Q10 (i) Factorise the following: $27y^3 + 125z^3$
Answer:
We know that
$a^3+b^3=(a+b)(a^2+b^2-ab)$
Now, we can write $27y^3 + 125z^3$ as
$\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3$
Here, $a = 3y \ \ and \ \ b = 5z$
Therefore,
$27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )$
$27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )$
Q10 (ii) Factorise the following: $64m^3 - 343n^3$
Answer:
We know that
$a^3-b^3=(a-b)(a^2+b^2+ab)$
Now, we can write $64m^3 - 343n^3$ as
$\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3$
Here, $a = 4m \ \ and \ \ b = 7n$
Therefore,
$64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )$
$64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )$
Q11 Factorise: $27x^3 + y^3 + z^3 - 9xyz$
Answer:
Given is $27x^3 + y^3 + z^3 - 9xyz$
Now, we know that
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Now, we can write $27x^3 + y^3 + z^3 - 9xyz$ as
$\Rightarrow 27x^3 + y^3 + z^3 - 9xyz$ $=(3x)^3+(y)^3+(z)^3-3.3x.y.z$
Here, $a= 3x , b = y \ \ and \ \ c = z$
Therefore,
$27x^3 + y^3 + z^3 - 9xyz$ $=(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )$
$=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )$
Answer:
We know that
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Now, multiply and divide the R.H.S. by 2
$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)$
$= \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)$
$= \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )$ $\left(\because a^2+b^2-2ab=(a-b)^2 \right )$
Hence proved.
Q13 If $x + y + z = 0$ , show that $x^3 + y^3 + z^3 = 3xyz$ .
Answer:
We know that
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Now, It is given that $x + y + z = 0$
Therefore,
$x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)$
$x^3+y^3+z^3-3xyz =0$
$x^3+y^3+z^3=3xyz$
Hence proved.
Answer:
Given is $(-12)^3 + (7)^3 + (5)^3$
We know that
If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$
Here, $x = -12 , y = 7 \ \ an d \ \ z = 5$
$\Rightarrow x+y+z = -12+7+5 = 0$
Therefore,
$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12)\times 7 \times 5 = -1260$
Therefore, value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$
Answer:
Given is $(28)^3 + (-15)^3 + (-13)^3$
We know that
If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$
Here, $x = 28 , y = -15 \ \ an d \ \ z = -13$
$\Rightarrow x+y+z =28-15-13 = 0$
Therefore,
$(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380$
Therefore, value of $(28)^3 + (-15)^3 + (-13)^3$ is $16380$
$25a^2 - 35a + 12$ |
Answer:
We know that
Area of rectangle is = $length \times breadth$
It is given that area = $25a^2-35a+12$
Now, by splitting middle term method
$\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12$
$= 5a(5a-4)-3(5a-4)$
$= (5a-3)(5a-4)$
Therefore, two answers are possible
case (i) :- Length = $(5a-4)$ and Breadth = $(5a-3)$
case (ii) :- Length = $(5a-3)$ and Breadth = $(5a-4)$
$35y^2 + 13y- 12$ |
Answer:
We know that
Area of rectangle is = $length \times breadth$
It is given that area = $35y^2 + 13y- 12$
Now, by splitting the middle term method
$\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12$
$= 7y(5y+4)-3(5y+4)$
$= (7y-3)(5y+4)$
Therefore, two answers are possible
case (i) :- Length = $(5y+4)$ and Breadth = $(7y-3)$
case (ii) :- Length = $(7y-3)$ and Breadth = $(5y+4)$
Q16 (i) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?
Volume : $3x^2 - 12x$ |
Answer:
We know that
Volume of cuboid is = $length \times breadth \times height$
It is given that volume = $3x^2-12x$
Now,
$\Rightarrow 3x^2-12x=3\times x\times (x-4)$
Therefore,one of the possible answer is possible
Length = $3$ and Breadth = $x$ and Height = $(x-4)$
Q16 (ii) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?
Volume : $12ky^2 + 8ky - 20k$ |
Answer:
We know that
Volume of cuboid is = $length \times breadth \times height$
It is given that volume = $12ky^2+8ky-20k$
Now,
$\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)$
$= k(12y^2+20y-12y-20)$
$= k\left(4y(3y+5)-4(3y+5) \right )$
$= k(3y+5)(4y-4)$
$= 4k(3y+5)(y-1)$
Therefore,one of the possible answer is possible
Length = $4k$ and Breadth = $(3y+5)$ and Height = $(y-1)$
Also Read:
In this exercise following topics are covered:
Also see-
An algebraic identity is a mathematical equation that holds true for all possible values of the variables in the equation.
Identity I : (x + y)²= x² + 2xy + y²
Identity II : (x – y)² = x² – 2xy + y²
Identity III : x² – y²= (x + y) (x – y)
Identity IV : (x + a) (x + b) = x²+ (a + b)x + ab
Identity V : (x + y + z)² = x²+ y²+ z² + 2xy + 2yz + 2zx
Identity VI : (x + y)³ = x³ + y³+ 3xy (x + y)
Identity VII : (x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³
Identity VIII : x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
(x + y)²= x² + 2xy + y²
(x – y)² = x² – 2xy + y²
x² – y²= (x + y) (x – y)
(x + y)³ = x³ + y³+ 3xy (x + y)
(x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³
(x + a) (x + b) = x² + (a + b)x + ab
(x + y + z)² = x²+ y²+ z² + 2xy + 2yz + 2zx
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
There are 16 questions in Class 9 Maths chapter 4 exercise 9. 5
There are 9 solved examples covered before NCERT solutions for Class 9 Maths chapter 4 exercise 9.5
Register for ALLEN Scholarship Test & get up to 90% Scholarship
Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters