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

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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.

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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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