RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions – Free PDF Download

RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions – Free PDF Download



Hey, are you a class 9 student and looking for ways to download RD Sharma Solutions for Class 9 Maths Chapter 5 “Factorization of Algebraic Expressions”? If yes. Then read this post till the end.

In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 5 in PDF that is prepared by Kota’s top IITian’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 9 Maths Chapter 5 “Factorization of Algebraic Expressions” in an easy way then you can use these solutions PDF.

Chapter 5 of RD Sharma Class 9 deals with Factorization of Algebraic Expressions using the factorization method. The process of factorization can be defined as the disintegration of a term into smaller factors whereas the algebraic expressions are built up of variables, integer constants, and basic arithmetic operations of algebra. If you want to improve your basic fundamentals of Factorization of Algebraic Expressions, then you can use this.

RD Sharma Solutions helps students to Practice important concepts of subjects easily. RD Sharma class 9 solutions provide detailed explanations of all the  exercise questions that students can use to clear their doubts instantly.

If you want to become good at Math then it is very important for you to have a good knowledge of all the important topics of class 9 math, so to learn and practice those topics you can use eSaral RD Sharma Solutions.

In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 5 that you can download to start your preparations anytime.

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RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions – Free PDF Download



Question 1: Factorize $x^{3}+x-3 x^{2}-3$

Solution: $x^{3}+x-3 x^{2}-3$

Here $x$ is common factor in $x^{3}+x$ and $-3$ is common factor in $-3 x^{2}-3$

$x^{3}-3 x^{2}+x-3$

$x^{2}(x-3)+1(x-3)$

Taking $(x-3)$ common

$(x-3)\left(x^{2}+1\right)$

Therefore $x^{3}+x-3 x^{2}-3=(x-3)\left(x^{2}+1\right)$

Question 2: Factorize $a(a+b)^{3}-3 a^{2} b(a+b)$

Solution: $a(a+b)^{3}-3 a^{2} b(a+b)$

Taking a $(a+b)$ as common factor

$=a(a+b)\left\{(a+b)^{2}-3 a b\right\}$

$=a(a+b)\left\{a^{2}+b^{2}+2 a b-3 a b\right\}$

$=a(a+b)\left(a^{2}+b^{2}-a b\right)$

Question 3: Factorize $x\left(x^{3}-y^{3}\right)+3 x y(x-y)$

Solution: $x\left(x^{3}-y^{3}\right)+3 x y(x-y)$

$=x(x-y)\left(x^{2}+x y+y^{2}\right)+3 x y(x-y)$

Taking $x(x-y)$ as a common factor

$=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$

$=x(x-y)\left(x^{2}+x y+y^{2}+3 y\right)$

Question 4: Factorize $\mathbf{a}^{2} \mathbf{x}^{2}+\left(\mathbf{a x}^{2}+1\right) \mathbf{x}+\mathbf{a}$

Solution: $a^{2} x^{2}+\left(a x^{2}+1\right) x+a$

$=a^{2} x^{2}+a+\left(a x^{2}+1\right) x$

$=a\left(a x^{2}+1\right)+x\left(a x^{2}+1\right)$

$=\left(a x^{2}+1\right)(a+x)$

Question 5: Factorize $x^{2}+y-x y-x$

Solution: $x^{2}+y-x y-x$

$=x^{2}-x-x y+y$

$=x(x-1)-y(x-1)$

$=(x-1)(x-y)$

Question 6: Factorize $x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}$

Solution: $x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}$

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

$=(x-2 y)\left(x^{2}+3 y^{2}\right)$

Question 7: Factorize $6 a b-b^{2}+12 a c-2 b c$

Solution: $6 a b-b^{2}+12 a c-2 b c$

$=6 a b+12 a c-b^{2}-2 b c$

Taking 6a common from first two terms and $-\mathrm{b}$ from last two terms

$=6 a(b+2 c)-b(b+2 c)$

Taking $(b+2 c)$ common factor

$=(b+2 c)(6 a-b)$

Question 8: Factorize $\left(x^{2}+1 / x^{2}\right)-4(x+1 / x)+6$

Solution: $\left(x^{2}+1 / x^{2}\right)-4(x+1 / x)+6$

$=x^{2}+1 / x^{2}-4 x-4 / x+4+2$

$=x^{2}+1 / x^{2}+4+2-4 / x-4 x$

$=\left(x^{2}\right)+(1 / x)^{2}+(-2)^{2}+2 x(1 / x)$

$+2(1 / x)(-2)+2(-2) x$

As we know, $x^{2}+y^{2}+z^{2}+2 x y$

$+2 y z+2 z x=(x+y+z)^{2}$

So, we can write;

$=(x+1 / x+(-2))^{2}$

or $(x+1 / x-2)^{2}$

Therefore, $\left.x^{2}+1 / x^{2}\right)-4(x+1 / x)+6$

$=(x+1 / x-2)^{2}$

Question 9: Factorize $x(x-2)(x-4)+4 x-8$

Solution: $x(x-2)(x-4)+4 x-8$

$=x(x-2)(x-4)+4(x-2)$

$=(x-2)[x(x-4)+4]$

$=(x-2)\left(x^{2}-4 x+4\right)$

$=(x-2)\left[x^{2}-2(x)(2)+(2)^{2}\right]$

$=(x-2)(x-2)^{2}$

$=(x-2)^{3}$

Question 10: Factorize $(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x$

Solution : $(x+2)\left(x^{2}+25\right)-10 x(x+2)$

Take $(x+2)$ as common factor;

$=(x+2)\left(x^{2}+25-10 x\right)$

$=(x+2)\left(x^{2}-10 x+25\right)$

Expanding the middle term of $\left(x^{2}-10 x+25\right)$

$=(x+2)\left(x^{2}-5 x-5 x+25\right)$

$=(x+2)\{x(x-5)-5(x-5)\}$

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

$=(x+2)(x-5)^{2}$

Therefore $(x+2)\left(x^{2}+25\right)-10 x(x+2)$

$=(x+2)(x-5)^{2}$

Question 11: Factorize $2 \mathbf{a}^{2}+2 \sqrt{6} \mathbf{a b}+\mathbf{3 b}^{2}$

Solution: $2 a^{2}+2 \sqrt{6} a b+3 b^{2}$

Above expression can be written as $(\sqrt{2 a})^{2}+2 \times \sqrt{2 a} \times \sqrt{3 b}+(\sqrt{3 b})^{2}$

As we know, $(p+q)^{2}=p^{2}+q^{2}+2 p q$

Here $p=\sqrt{2} a$ and $q=\sqrt{3} b$

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

Therefore, $2 a^{2}+2 \sqrt{6} a b+3 b^{2}=(\sqrt{2} a+\sqrt{3 b})^{2}$

Question 12: Factorize $(a-b+c)^{2}+(b-c+a)^{2}+2(a-b+c)(b-c+a)$

Solution: $(a-b+c)^{2}+(b-c+a)^{2}$

$+2(a-b+c)(b-c+a)$

$\left\{\right.$ Because $\left.p^{2}+q^{2}+2 p q=(p+q)^{2}\right\}$

Here $p=a-b+c$ and $q=b-c+a$

$=[a-b+c+b-c+a]^{2}$

$=(2 a)^{2}$

$=4 a^{2}$

Question 13: Factorize $a^{2}+b^{2}+2(a b+b c+c a)$

Solution: $a^{2}+b^{2}+2 a b+2 b c+2 c a$

As we know, $p^{2}+q^{2}+2 p q=(p+q)^{2}$

We get,

$=(a+b)^{2}+2 b c+2 c a$

$=(a+b)^{2}+2 c(b+a)$

Or $(a+b)^{2}+2 c(a+b)$

Take $(a+b)$ as common factor;

$=(a+b)(a+b+2 c)$

Therefore, $a^{2}+b^{2}+2 a b+2 b c+2 c a$

$=(a+b)(a+b+2 c)$

Question 14: Factorize $4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}$

Solution : Consider $(x-y)=p,(x+y)=q$

$=4 p^{2}-12 p q+9 q^{2}$

Expanding the middle term, $-12=-6-6$ also $4 \times 9=-6 \times-6$

$=4 p^{2}-6 p q-6 p q+9 q^{2}$

$=2 p(2 p-3 q)-3 q(2 p-3 q)$

$=(2 p-3 q)(2 p-3 q)$

$=(2 p-3 q)^{2}$

Substituting back $p=x-y$ and $q=x+y;$

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

$=[2 x-2 y-3 x-3 y]^{2}$

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

$=[-x-5 y]^{2}$

$=[(-1)(x+5 y)]^{2}$

$=(x+5 y)^{2}$

Therefore, $4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}$

$=(x+5 y)^{2}$

Question 15: Factorize $\mathbf{a}^{2}-\mathbf{b}^{2}+\mathbf{2 b c}-\mathbf{c}^{2}$

Solution : $a^{2}-b^{2}+2 b c-c^{2}$

As we know, $(a-b)^{2}=a^{2}+b^{2}-2 a b$

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

Also we know, $a^{2}-b^{2}=(a+b)(a-b)$

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

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

Therefore, $a^{2}-b^{2}+2 b c-c^{2}$

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

Question 16: Factorize $\mathbf{a}^{2}+2 \mathrm{ab}+\mathbf{b}^{2}-\mathbf{c}^{2}$

Solution: $a^{2}+2 a b+b^{2}-c^{2}$

$=\left(a^{2}+2 a b+b^{2}\right)-c^{2}$

$=(a+b)^{2}-(c)^{2}$

We know, $a^{2}-b^{2}=(a+b)(a-b)$

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

Therefore $a^{2}+2 a b+b^{2}-c^{2}$

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

Exercise $5.2$ Page No: $5.13$

Factorize each of the following expressions:

Question 1: $\mathbf{p}^{3}+\mathbf{2 7}$

Solution: $p^{3}+27$

$=p^{3}+3^{3}$

$\left[\right.$ using $\left.a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(p+3)\left(p^{2}-3 p-9\right)$

Therefore, $p^{3}+27$

$(p+3)\left(p^{2}-3 p-9\right)$

Question 2: $\mathrm{y}^{3}+125$

Solution: $y^{3}+125$

$=y^{3}+5^{3}$

$\left[u \operatorname{sing} a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(y+5)\left(y^{2}-5 y+5^{2}\right)$

$=(y+5)\left(y^{2}-5 y+25\right)$

Therefore, $y^{3}+125=(y+5)\left(y^{2}-5 y+25\right)$

Question $3: 1-27 a^{3}$

Solution: $=(1)^{3}-(3 a)^{3}$

[using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=(1-3 a)\left(1^{2}+1 \times 3 a+(3 a)^{2}\right)$

$=(1-3 a)\left(1+3 a+9 a^{2}\right)$

Therefore, $1-27 \mathrm{a}^{3}=(1-3 \mathrm{a})\left(1+3 \mathrm{a}+9 \mathrm{a}^{2}\right)$

Question 4: $8 x^{3} y^{3}+27 a^{3}$

Solution: $8 x^{3} y^{3}+27 a^{3}$

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

$\left[\right.$ using $\left.a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(2 x y+3 a)\left((2 x y)^{2}-2 x y \times 3 a+(3 a)^{2}\right)$

$=(2 x y+3 a)\left(4 x^{2} y^{2}-6 x y a+9 a^{2}\right)$

Question 5: $64 \mathbf{a}^{3}-\mathbf{b}^{3}$

Solution: $64 a^{3}-b^{3}$

$=(4 a)^{3}-b^{3}$

$\left[\right.$ using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=(4 a-b)\left((4 a)^{2}+4 a \times b+b^{2}\right)$

$=(4 a-b)\left(16 a^{2}+4 a b+b^{2}\right)$

Question $6: x^{3} / 216-8 y^{3}$

Solution: $x^{3} / 216-8 y^{3}$

$=\left(\frac{x}{6}\right)^{3}-(2 y)^{3}$

$\because\left[x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$

$=\left(\frac{x}{6}-2 y\right)\left(\left(\frac{x}{6}\right)^{2}+\frac{x}{6} \times 2 y+(2 y)^{2}\right)$

$=\left(\frac{x}{6}-2 y\right)\left(\frac{x^{2}}{36}+\frac{x y}{3}+4 y^{2}\right)$

$\therefore \frac{x^{3}}{216}-8 y^{3}=\left(\frac{x}{6}-2 y\right)$

$\left(\frac{x^{2}}{36}+\frac{x y}{3}+4 y^{2}\right)$

Question 7: $10 x^{4} y-10 x y^{4}$

Solution: $10 x^{4} y-10 x y^{4}$

$=10 x y\left(x^{3}-y^{3}\right)$

$\left[\right.$ using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=10 x y(x-y)\left(x^{2}+x y+y^{2}\right)$

Therefore, $10 x^{4} y-10 x y^{4}$

$=10 x y(x-y)\left(x^{2}+x y+y^{2}\right)$

Question 8: $54 x^{6} y+2 x^{3} y^{4}$

Solution: $54 x^{6} y+2 x^{3} y^{4}$

$=2 x^{3} y\left(27 x^{3}+y^{3}\right)$

$=2 x^{3} y\left((3 x)^{3}+y^{3}\right)$

$\left[u \operatorname{sing} a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=2 x^{3} y\left\{(3 x+y)\left((3 x)^{2}-3 x y+y^{2}\right)\right\}$

$=2 x^{3} y(3 x+y)\left(9 x^{2}-3 x y+y^{2}\right)$

Question $9: 32 \mathrm{a}^{3}+108 b^{3}$

Solution: $32 a^{3}+108 b^{3}$

$=4\left(8 a^{3}+27 b^{3}\right)$

$=4\left((2 a)^{3}+(3 b)^{3}\right)$

$\left[\right.$ using $\left.a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=4\left[(2 a+3 b)\left((2 a)^{2}-2 a \times 3 b+(3 b)^{2}\right)\right]$

$=4(2 a+3 b)\left(4 a^{2}-6 a b+9 b^{2}\right)$

Question $10:(a-2 b)^{3}-512 b^{3}$

Solution: $(a-2 b)^{3}-512 b^{3}$

$=(a-2 b)^{3}-(8 b)^{3}$

$\left[\right.$ using $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=(a-2 b-8 b)\left\{(a-2 b)^{2}+(a-2 b) 8 b+(8 b)^{2}\right\}$

$=(a-10 b)\left(a^{2}+4 b^{2}-4 a b+8 a b-16 b^{2}+64 b^{2}\right)$

$=(a-10 b)\left(a^{2}+52 b^{2}+4 a b\right)$

Question 11: $(a+b)^{3}-8(a-b)^{3}$

Solution: $(a+b)^{3}-8(a-b)^{3}$

$=(a+b)^{3}-[2(a-b)]^{3}$

$=(a+b)^{3}-[2 a-2 b]^{3}$

$\left[\right.$ using $\left.p^{3}-q^{3}=(p-q)\left(p^{2}+p q+q^{2}\right)\right]$

Here $p=a+b$ and $q=2 a-2 b$

$=(a+b-(2 a-2 b))\left((a+b)^{2}+(a+b)(2 a-2 b)+(2 a-2 b)^{2}\right)$

$=(a+b-2 a+2 b)\left(a^{2}+b^{2}+2 a b+(a+b)\right.$

$\left.(2 a-2 b)+(2 a-2 b)^{2}\right)$

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

$\left(a^{2}+b^{2}+2 a b+2 a^{2}-2 a b+2 a b-2 b^{2}+(2 a-2 b)^{2}\right)$

$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+(2 a-2 b)^{2}\right)$

$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+4 a^{2}+4 b^{2}-8 a b\right)$

$=(3 b-a)\left(3 a^{2}+4 a^{2}-b^{2}+4 b^{2}-8 a b+2 a b\right)$

$=(3 b-a)\left(7 a^{2}+3 b^{2}-6 a b\right)$

Question 12: $(x+2)^{3}+(x-2)^{3}$

Solution: $(x+2)^{3}+(x-2)^{3}$

$\left[\right.$ using $\left.p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right)\right]$

Here $p=x+2$ and $q=x-2$

$=(x+2+x-2)\left((x+2)^{2}-(x+2)(x-2)+(x-2)^{2}\right)$

$=2 x\left(x^{2}+4 x+4-(x+2)(x-2)+x^{2}-4 x+4\right)$

$\left[\right.$ Using $\left.:(a+b)(a-b)=a^{2}-b^{2}\right]$

$=2 x\left(2 x^{2}+8-\left(x^{2}-2^{2}\right)\right)$

$=2 x\left(2 x^{2}+8-x^{2}+4\right)$

$=2 x\left(x^{2}+12\right)$

Exercise $5.3$ Page No: $5.17$

Question 1: Factorize $64 \mathrm{a}^{3}+125 \mathrm{~b}^{3}+240 \mathrm{a}^{2} \mathrm{~b}+300 \mathrm{ab}^{2}$

Solution: $64 a^{3}+125 b^{3}+240 a^{2} b+300 a b^{2}$

$=(4 a)^{3}+(5 b)^{3}+3(4 a)^{2}(5 b)+3(4 a)(5 b)^{2}$,

which is similar to $a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

We know that, $\left.a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$

$=(4 a+5 b)^{3}$

Question 2: Factorize $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$

Solution: $125 x^{3}-27 y^{3}-225 x^{2} y+135 x y^{2}$

Above expression can be written as $(5 x)^{3}-(3 y)^{3}-3(5 x)^{2}(3 y)+3(5 x)(3 y)^{2}$

Using: $a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}$

$=(5 x-3 y)^{3}$

Question 3: Factorize $8 / 27 \mathrm{x}^{3}+1+4 / 3 \mathrm{x}^{2}+2 \mathrm{x}$

Solution: $8 / 27 x^{3}+1+4 / 3 x^{2}+2 x$

$=\left(\frac{2}{3} x^{3}\right)^{3}+1^{3}+3 \times\left(\frac{2}{3} x\right)^{2}$

$\times 1+3(1)^{2} \times\left(\frac{2}{3} x\right)$

$\left[\because x^{3}+b^{3}+3 x^{2} b+3 x b^{2}=(x+b)^{3}\right]$

$\therefore \frac{8}{27} x^{3}+1+\frac{4}{3} x^{2}+2 x$

$=\left(\frac{2}{3} x+1\right)^{3}$

Question 4: Factorize $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$

Solution: $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$

Above expression can be written as $(2 x)^{3}+(3 y)^{3}+3 x(2 x)^{2} \times 3 y+3 x(2 x)(3 y)^{2}$

Which is similar to $\left.a^{3}+b^{3}+3 a^{2} b+3 a b^{2}=(a+b)^{3}\right]$

Here $\mathrm{a}=2 \mathrm{x}$ and $\mathrm{b}=3 \mathrm{y}$

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

Therefore, $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}=(2 x+3 y)^{3}$

Question 5: Factorize $a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$

Solution: $a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$

Using: $a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}$

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

Again, Using: $\left.a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(a-b+2)\left((a-b)^{2}-(a-b) \times 2+2^{2}\right)$

$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2(a-b)+4\right)$

$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$

$a^{3}-3 a^{2} b+3 a b^{2}-b^{3}+8$

$=(a-b+2)\left(a^{2}+b^{2}-2 a b-2 a+2 b+4\right)$

Exercise $5.4$ Page No: $5.22$

Factorize each of the following expressions:

Question $1: \mathrm{a}^{3}+8 \mathrm{~b}^{3}+64 \mathrm{c}^{3}-24 \mathrm{abc}$

Solution: $a^{3}+8 b^{3}+64 c^{3}-24 a b c$

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

$\left[\right.$ Using $\left.a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

$=(a+2 b+4 c)\left(a^{2}+(2 b)^{2}+(4 c)^{2}-a \times 2 b-2 b \times 4 c-4 c \times a\right)$

$=(a+2 b+4 c)\left(a^{2}+4 b^{2}+16 c^{2}-2 a b-8 b c-4 a c\right)$

Therefore, $a^{3}+8 b^{3}+64 c^{3}-24 a b c=(a+2 b+4 c)$

$\left(a^{2}+4 b^{2}+16 c^{2}-2 a b-8 b c-4 a c\right)$

Question 2: $x^{3}-8 y^{3}+27 z^{3}+18 x y z$

Solution: $=x^{3}-(2 y)^{3}+(3 z)^{3}-3 \times x \times(-2 y)(3 z)$

$=(x+(-2 y)+3 z)\left(x^{2}+(-2 y)^{2}+(3 z)^{2}\right.$

$-x(-2 y)-(-2 y)(3 z)-3 z(x))$

$\left[\right.$ using $a^{3}+b^{3}+c^{3}-3 a b c$

$\left.=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

$=(x-2 y+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 z x\right)$

Question $3: 27 x^{3}-y^{3}-z^{3}-9 x y z$

Solution: $27 x^{3}-y^{3}-z^{3}-9 x y z$

$=(3 x)^{3}-y^{3}-z^{3}-3(3 x y z)$

$\left[\mathrm{U} \sin g a^{3}+b^{3}+c^{3}-3 a b c\right.$

$\left.=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

Here $a=3 x, b=-y$ and $c=-z$

$=(3 x-y-z)\left\{(3 x)^{2}+(-y)^{2}\right.$

$\left.\left.+(-z)^{2}+3 x y-y z+3 x z\right)\right\}$

$=(3 x-y-z)\left\{9 x^{2}+y^{2}+z^{2}\right.$

$+3 x y-y z+3 x z)\}$

Question 4: $1 / 27 \mathrm{x}^{3}-\mathrm{y}^{3}+125 \mathrm{z}^{3}+5 \mathrm{xyz}$

Solution: $1 / 27 x^{3}-y^{3}+125 z^{3}+5 x y z$

$=(x / 3)^{3}+(-y)^{3}+(5 z)^{3}-3 x / 3(-y)(5 z)$

$\left[\right.$ Using $\left.a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

$=(x / 3+(-y)+5 z)\left((x / 3)^{2}+(-y)^{2}+(5 z)^{2}-x / 3(-y)-(-y) 5 z-5 z(x / 3)\right)$

$=(x / 3-y+5 z)\left(x^{\wedge} 2 / 9+y^{2}+25 z^{2}+x y / 3+5 y z-5 z x / 3\right)$

Question 5: $8 x^{3}+27 y^{3}-216 z^{3}+108 x y z$

Solution: $8 x^{3}+27 y^{3}-216 z^{3}+108 x y z$

$=(2 x)^{3}+(3 y)^{3}+(-6 y)^{3}-3(2 x)(3 y)(-6 z)$

$=(2 x+3 y+(-6 z))\left\{(2 x)^{2}+(3 y)^{2}\right.$

$\left.+(-6 z)^{2}-2 x \times 3 y-3 y(-6 z)-(-6 z) 2 x\right\}$

$=(2 x+3 y-6 z)\left\{4 x^{2}+9 y^{2}+36 z^{2}-6 x y+18 y z+12 z x\right\}$

Question $6: 125+8 x^{3}-27 y^{3}+90 x y$

Solution: $125+8 x^{3}-27 y^{3}+90 x y$

$=(5)^{3}+(2 x)^{3}+(-3 y)^{3}-3 \times 5 \times 2 x \times(-3 y)$

$=(5+2 x+(-3 y))\left(5^{2}+(2 x)^{2}+(-3 y)^{2}\right.$

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

$=(5+2 x-3 y)\left(25+4 x^{2}+9 y^{2}-10 x+6 x y+15 y\right)$

Question $7:(3 x-2 y)^{3}+(2 y-4 z)^{3}+(4 z-3 x)^{3}$

Solution: $(3 x-2 y)^{3}+(2 y-4 z)^{3}+(4 z-3 x)^{3}$

Let $(3 x-2 y)=a,(2 y-4 z)=b,(4 z-3 x)=c$

$a+b+c=3 x-2 y+2 y-4 z+4 z-3 x=0$

We $k n o w, a^{3}+b^{3}+c^{3}-3 a b c$

$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$\Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=0$

or $a^{3}+b^{3}+c^{3}=3 a b c$

$\Rightarrow(3 x-2 y)^{3}+(2 y-4 z)^{3}+(4 z-3 x)^{3}$

$=3(3 x-2 y)(2 y-4 z)(4 z-3 x)$

Question 8: $(2 x-3 y)^{3}+(4 z-2 x)^{3}+(3 y-4 z)^{3}$

Solution: $(2 x-3 y)^{3}+(4 z-2 x)^{3}+(3 y-4 z)^{3}$

Let $2 x-3 y=a, 4 z-2 x=b, 3 y-4 z=c$

$a+b+c=2 x-3 y+4 z-2 x+3 y-4 z=0$

We $\mathrm{know}, \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}-3 \mathrm{abc}=$

$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$\Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=0$

$(2 x-3 y)^{3}+(4 z-2 x)^{3}+(3 y-4 z)^{3}$

$=3(2 x-3 y)(4 z-2 x)(3 y-4 z)$

Exercise VSAQs Page No: $5.24$

Question 1: Factorize $x^{4}+\mathbf{x}^{2}+25$

Solution: $x^{4}+x^{2}+25$

$=\left(x^{2}\right)^{2}+5^{2}+x^{2}$

$\left[\right.$ using $\left.a^{2}+b^{2}=(a+b)^{2}-2 a b\right]$

$=\left(x^{2}+5\right)^{2}-2\left(x^{2}\right)(5)+x^{2}$

$=\left(x^{2}+5\right)^{2}-10 x^{2}+x^{2}$

$=\left(x^{2}+5\right)^{2}-9 x^{2}$

$=\left(x^{2}+5\right)^{2}-(3 x)^{2}$

$\left[u \operatorname{sing} a^{2}-b^{2}=(a+b)(a-b]\right.$

$=\left(x^{2}+3 x+5\right)\left(x^{2}-3 x+5\right)$

Question 2: Factorize $x^{2}-1-2 a-a^{2}$

Solution: $x^{2}-1-2 a-a^{2}$

$x^{2}-\left(1+2 a+a^{2}\right)$

$x^{2}-(a+1)^{2}$

$(x-(a+1)(x+(a+1)$

$(x-a-1)(x+a+1)$

$\left[\right.$ using $a^{2}-b^{2}=(a+b)(a-b)$ and $\left.(a+b)^{\wedge} 2=a^{\wedge} 2+b^{\wedge} 2+2 a b\right]$

Question $3:$ If $a+b+c=0$, then write the value of $a^{3}+b^{3}+c^{3}$.

Solution: We know, $a^{3}+b^{3}+c^{3}-3 a b c$

$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

Put $a+b+c=0$

This implies

$a^{3}+b^{3}+c^{3}=3 a b c$

Question 4: If $a^{2}+b^{2}+c^{2}=20$ and $a+b+c=0$, find $a b+b c+c a$

Solution: We know, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$

$0=20+2(a b+b c+c a)$

$-10=a b+b c+c a$

Or $a b+b c+c a=-10$

Question 5: If $a+b+c=9$ and $a b+b c+c a=40$, find $a^{2}+b^{2}+c^{2}$

Solution: We know, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$

$9^{2}=a^{2}+b^{2}+c^{2}+2(40)$

$81=a^{2}+b^{2}+c^{2}+80$

$\Rightarrow a^{2}+b^{2}+c^{2}=1$

Also Read,

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