# Exercise 2.5 Polynomials NCERT Solutions Class 9

Go back to  'Polynomials'

## Chapter 2 Ex.2.5 Question 1

Use suitable identities to find the following products:

(i) \begin{align}(x+4)(x+10)\end{align}

(ii) \begin{align}(x+8)(x-10)\end{align}

(iii) \begin{align}(3 x+4)(3 x-5)\end{align}

(iv) \begin{align}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\end{align}

(v) \begin{align}(3-2 x)(3+2 x)\end{align}

### Solution

​​​​​​Reasoning:

Identities: \begin{align}(x+a)(x+b)&=x^{2}+(a+b) x+a b \\ (a+b)(a-b)&=a^{2}-b^{2}\end{align}

Steps:

\begin{align}\text{(i)}\;\;(x+4)(x+10)\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

Here \begin{align}\text{a} = {4}, \text{b} = 10\end{align}

\begin{align}&(x+4)(x+10) \\&=x^{2}+(4+10) x+4 \times 10 \\ &=x^{2}+14 x+40 \end{align}

\begin{align}\text{(ii)}\;\;(x+8)(x-10)\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

Here $$a =8, b=-10$$

\begin{align}&(x+8)(x-10) \\&=x^{2}+(8-10) x+(8)(-10) \\ &=x^{2}-2 x-80 \end{align}

\begin{align}\text{(iii)}\;\;(3 x+4)(3 x-5)\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

Here \begin{align}x \rightarrow 3 x, a=4, b=-5\end{align}

\begin{align}&(3 x+4)(3 x-5) \\&=(3 x)^{2}+(4-5)(3 x)+(4)(-5) \\ &=9 x^{2}-3 x-20 \end{align}

(iv) \begin{align}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\end{align}

Identity: \begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}

Here \begin{align}a=y^{2}, b=\frac{3}{2}\end{align}

\begin{align}&\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) \\&=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} \\ &=y^{4}-\frac{9}{4} \end{align}

(v) \begin{align}(3-2 x)(3+2 x) \end{align}

Identity: \begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}

Here \begin{align}a=3, b=2 x\end{align}

\begin{align}&(3-2 x)(3+2 x) \\&=(3)^{2}-(2 x)^{2} \\ &=9-4 x^{2} \end{align}

## Chapter 2 Ex.2.5 Question 2

Evaluate the following products without multiplying directly:

(i) \begin{align}103 \times 107\end{align}

(ii)  \begin{align}95 \times 96\end{align}

(iii)  \begin{align}104 \times 96\end{align}

### Solution

Reasoning:

Identities:

\begin{align}(x+a)(x+b)&=x^{2}+(a+b) x+a b \\ (a+b)(a-b)&=a^{2}-b^{2}\end{align}

Steps:

(i) \begin{align}103 \times 107\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

\begin{align} 103 \times 107 &=(100+3)(100+7) \\ &=(100)^{2}\!\!+\!(3\!+\!7)\!(100)\!+\!\!(3)(7) \\\\text {Taking } x&=100, a=3, b=7 ) \\ &=10000+1000+21 \\ &=11021 \end{align} (ii) \(\begin{align}95 \times 96\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

\begin{align}{95 \times 96}&={ (100\!-\!5)(100\!-\!4)}\\&={\left[\!\begin{array}{l} {(100)^2}\!\!+\!(\!- 5\!- 4)\\(100)\!+\!( - 5)( - 4)\end{array}\!\right]}\\\\{\rm{(Taking }\;x \!=\! 100,a}&{ = - 5,b = - 4)}\\&{ = 10000 - 900 + 20}\\&{ = 9120}\end{align}

(iii)  \begin{align}104 \times 96\end{align}

Identity: \begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}

\begin{align} 104 \times 96 &=(100+4)(100-4) \\ &=(100)^{2}-(4)^{2} \\\\text { Taking } a&=100, b=4 ) \\ &=10000-16 \\ &=9984 \end{align} ## Chapter 2 Ex.2.5 Question 3 Factorise the following using appropriate identities: (i) \(\begin{align}9 x^{2}+6 x y+y^{2}\end{align}

(ii) \begin{align} 4 y^{2}-4 y+1\end{align}

(iii) \begin{align} x^{2}-\frac{y^{2}}{100}\end{align}

### Solution

Reasoning:

Identities:

\begin{align}& {(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {} & {(a-b)^{2}=a^{2}-2 a b+b^{2}} \\ {} & {(a+b)(a-b)=a^{2}-b^{2}}\end{align}

Steps:

(i) \begin{align}9 x^{2}+6 x y+y^{2}\!=\!(3 x)^{2}\!+\!2(3 x)(y)\!+\!(y)^{2}\end{align}

Identity:  \begin{align}(a+b)^{2}=a^{2}+2 a b+b^{2}\end{align}

Here \begin{align} a=3 x, b=y \end{align}

Hence   \begin{align}9 x^{2}+6 x y+y^{2}=(3 x+y)^{2}\end{align}

(ii) \begin{align}4 y^{2}-4 y+1=\left(2 y^{2}\right)-2(2 y)(1)+(1)^{2}\end{align}

Identity: \begin{align}(a-b)^{2}=a^{2}-2 a b+b^{2}\end{align}

Here \begin{align} a=2 y, b=1\end{align}

Hence \begin{align} 4 y^{2}-4 y+1=(2 y-1)^{2}\end{align}

(iii) \begin{align}x^{2}-\frac{y^{2}}{100}=(x)^{2}-\left(\frac{y}{10}\right)^{2}\end{align}

Identity: \begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}

Here \begin{align}&{ a=x, b=\frac{y}{10}} \end{align}

Hence \begin{align}{ x^{2}-\frac{y^{2}}{100}=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)}\end{align}

## Chapter 2 Ex.2.5 Question 4

Expand each of the following, using suitable identities:

(i) \begin{align}(x+2 y+4 z)^{2}\end{align}

(ii) \begin{align}(2 x-y+z)^{2}\end{align}

(iii)\begin{align}(-2 x+3 y+2 z)^{2}\end{align}

(iv) \begin{align}(3 a-7 b-c)^{2}\end{align}

(v)\begin{align}(-2 x+5 y-3 z)^{2}\end{align}

(vi) \begin{align}\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}\end{align}

### Solution

Reasoning:

Identity:\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Steps:

\begin{align}(\mathrm{i})\;\;(x+2 y+4 z)^{2}\end{align}

Identity:\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=x, b=2 y, c=4 z\end{align}

\begin{align}&{(x + 2y + 4z)^2}&\\&=\left[ \begin{array}{l}{x^2} + {(2y)^2} + {(4z)^2} + 2(x)(2y) + \\2(2y)(4z) + 2(4z)(x)\end{array} \right]\\ &= {x^2} + 4{y^2} + 16{z^2} + 4xy + 16yz + 8zx\end{align}

\begin{align}\text {(ii)}\left(2 x-y+z)^{2}\right.\end{align}

Identity: \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=2 x, b=-y, c=z\end{align}

\begin{align}&{(2x - y + z)^2}\\ &= \left[ {\begin{array}{*{20}{l}}\begin{array}{l}{(2x)^2} + {( - y)^2} + {(z)^2}\\ + 2(x)( - y) + \end{array}\\{2( - y)(z) + 2(z)(2x)}\end{array}} \right]\\&= 4{x^2} + {y^2} + {z^2} + 4xy + 2yz + 4zx\end{align}

\begin{align}\text { (iii) }\;\;(-2 x+3 y+2 z)^{2}\end{align}

Identity: \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+\\c^{2}+2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=2 x, b=3 y, c=2 z\end{align}

\begin{align}&{( - 2x + 3y + 2z)^2}\\ &= \left[ {\begin{array}{*{20}{l}}\begin{array}{l}{( - 2x)^2} + {(3y)^2} + {(2z)^2}\\ + 2( - 2x)(3y) + \end{array}\\{2(3y)(2z) + 2(2z)(2x)}\end{array}} \right]\\ &= 4{x^2} + 9{y^2} + 4{z^2} - 12xy + 12yz - 8zx\end{align}

\begin{align}(\text{iv})(3 a-7 b-c)^{2}\end{align}

Identity: \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=3 a, b=-7 b, c=-c\end{align}

\begin{align}&{(3a - 7b - c)^2}\\ &= \left[ \begin{array}{l}{(3a)^2} + {( - 7b)^2} + {( - c)^2}\\ + 2(3a)( - 7b)\\ + 2( - 7b)( - c) + 2( - c)(3a)\end{array} \right]\\ &= 9{a^2} + 49{b^2} + {c^2} - 42ab + 14bc - 6ca\end{align}

\begin{align}(\mathrm{v})\;\;(-2 x+5 y-3 z)^{2}\end{align}

Identity: \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=-2 x, b=5 y, c=-3 z\end{align}

\begin{align}&{( - 2x + 5y - 3z)^2}\\ &= \left[ \begin{array}{l}{( - 2x)^2} + {(5y)^2} + {( - 3z)^2} +\\ 2( - 2x)(5y)\\ + 2(5y)( - 3z) + 2( - 3z)( - 2x)\end{array} \right]\\& = \!4{x^2}\!\!+\! 25{y^2} + 9{z^2} \!\!-\! 20xy - 30yz \!+\!\! 12zx\end{align}

\begin{align}(\mathrm{vi})\;\;\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2} \end{align}

Identity: \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}

Taking \begin{align}a=\frac{1}{4} a, b=\frac{-1}{2} b, c=1 \end{align}

\begin{align}&{\left( {\frac{1}{4}a - \frac{1}{2}b + 1} \right)^2}\\&= \left[ \begin{array}{l}{\left( {\frac{1}{4}a} \right)^2} + {\left( {\frac{{ - 1}}{2}b} \right)^2} + {(1)^2} + \\2\left( {\frac{1}{4}a} \right)\left( {\frac{{ - 1}}{2}b} \right) + \\2\left( {\frac{{ - 1}}{2}b} \right)(1) + 2(1)\left( {\frac{1}{4}a} \right)\end{array} \right]\\&= \frac{1}{{16}}{a^2} + \frac{1}{4}{b^2} + 1 - \frac{1}{4}ab - b + \frac{1}{2}a\end{align}

## Chapter 2 Ex.2.5 Question 5

Factorise:

(i)\begin{align}\left[ \begin{array}{l} 4 x^{2}+9 y^{2}+16 z^{2}+\\12 x y-24 y z-16 x z\end{array} \right]\end{align}

(ii)\begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+\\4 \sqrt{2} y z-8 x z\end{array} \right]\end{align}

### Solution

Reasoning:

Identity:  \begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+\\c^{2}+2 a b+2 b c+2 c a\end{array} \right]\end{align}

Steps:

(i) \begin{align}\left[ \begin{array}{l} 4 x^{2}+9 y^{2}+16 z^{2}+\\12 x y-24 y z-16 x z\end{array} \right]\end{align}

This can be re-written as:

$\left[ \begin{array}{l}\left( {2{x^2}} \right) + {(3y)^2} + \left( { - 4{z^2}} \right) +\\ 2(2x)(3y) + 2(3y)( - 4z) +\\ 2( - 4z)(2x) + 2(2x)( - 4z)\end{array} \right]$

Which is of the form: \begin{align}\left[ \begin{array}{l}a^{2}+b^{2}+c^{2}+2 a b+2 b c+\\2 c a=(a+b+c)^{2}\end{array} \right]\end{align}

Here \begin{align}a=2 x , b=3 y, c=-4 z\end{align}

Hence \begin{align}\left[ \begin{array}{l}4 x^{2}+9 y^{2}+16 z^{2}+12 x y-\\24 y z-16 x z=(2 x+3 y-4 z)^{2}\end{array} \right]\end{align}

(ii) \begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+\\4 \sqrt{2} y z-8 x z\end{array} \right]\end{align}

This can be re-written as:

$\left[ \begin{array}{l}{( - \sqrt 2 x)^2} + {(y)^2} + {(2\sqrt 2 z)^2} + \\2( - \sqrt 2 x)(y) + 2(y)(2\sqrt 2 z) +\\ 2(2\sqrt 2 z)( - \sqrt 2 x)\end{array} \right]$

Which is of the form: \begin{align}\left[ \begin{array}{l}a^{2}+b^{2}+c^{2}+2 a b+2 b c+\\2 c a=(a+b+c)^{2}\end{array} \right]\end{align}

Here \begin{align}a=-2 \sqrt{2} x, b=y, c=2 \sqrt{2} z\end{align}

Hence  \begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-\\8 x z=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}\end{array} \right]\end{align}

## Chapter 2 Ex.2.5 Question 6

Write the following cubes in expanded form:

(i) \begin{align}(2 x+1)^{3} \end{align}

(ii) \begin{align}(2 a-3 b)^{3}\end{align}

(iii) \begin{align}\left(\frac{3}{2} x+1\right)^{3} \end{align}

(iv) \begin{align}\left(x-\frac{2}{3} y\right)^{3}\end{align}

### Solution

Reasoning:

Identities:

\begin{align} &(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y) \\ &(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y) \end{align}

Steps:

(i) \begin{align}(2 x+1)^{3}\end{align}

Identity: \begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}

Here \begin{align}x=2 x\;,\; y=1\end{align}

\begin{align}&(2 x+1)^{3}\\&=(2 x)^{3}\!\!+\!(1)^{3}\!\!+\!3(2 x)\!(1)\!(2 x\!+\!1) \\ &=8 x^{3}+1+6 x(2 x+1) \\ &=8 x^{3}+1+12 x^{2}+6 x \\ &=8 x^{3}+12 x^{2}+6 x+1 \end{align}

(ii) \begin{align}\left(2 a-3 b)^{3}\right.\end{align}

Identity: \begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}

Here $$x = 2a, y = 3b$$

\begin{align}&(2 a-3 b)^{3} \\&\!=\!(2 a)^{3}\!\!-\!(3 b)^{3}\!-\!3(2 a)(3 b)(2 a\!-\!3 b) \\ &=8 a^{3}-27 b^{3}-18 a b(2 a-3 b) \\ &=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2} \\ &=8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} \end{align}

(iii) \begin{align}\left[\frac{3}{2} x+1\right]^{3} \end{align}

Identity: \begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}

Here \begin{align}x=\frac{3}{2}, y=1\end{align}

\begin{align}&\left(\frac{3}{2} x+1\right)^{3} \\&=\left(\frac{3}{2}x\!\!\right)^{3}+\!(1)^{3}\!+\!3\left(\frac{3}{2} x\right)\!\!(1)\!\!\left(\frac{3}{2} x\!+\!1\!\!\right) \\ &=\frac{27}{8} x^{3}+1+\frac{9}{2} x+\left(\frac{3}{2} x+1\right) \\ &=\frac{27}{8} x^{3}+1+\frac{27}{8} x^{2}+\frac{9}{2} x \\ &=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1 \end{align}

(iv) \begin{align}\left(x-\frac{2}{3} y\right)^{3} \end{align}

Identity: \begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}

Here \begin{align}x=x, y=\frac{2}{3} y\end{align}

\begin{align}&\left(x-\frac{2}{3} y\right)^{3} \\&=\!x^{3}\!-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right) \\ &=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right) \\ &=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2} \\ &=x^{3}-2 x^{2} y+\frac{4}{3} x y^{2}-\frac{8}{27} y^{3} \end{align}

## Chapter 2 Ex.2.5 Question 7

Evaluate the following using suitable identities:

(i) \begin{align}(99)^{3}\end{align}

(ii) \begin{align}(102)^{3} \end{align}

(iii) \begin{align}(998)^{3} \end{align}

### Solution

Reasoning:

Identities:

\begin{align} (x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\\{(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)}\end{align}

Steps:

(i)\begin{align}\;\;(99)^{3}=(100-1)^{3}\end{align}

Identity: \begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}

Take \begin{align}x=100, y=1\end{align}

\begin{align}(99)^{3} &\!=\!(100)^{3}\!\!-\!(1)^{3}\!\!-\!3(100)\!(1)\!(100\!-\!\!1) \\ &=1000000-1-300 \times 99 \\ &=999999-29700 \\ &=9,70,299 \end{align}

(ii)\begin{align}\;\;(102)^{3}=(100+2)^{3}\end{align}

Identity:  \begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}

Take \begin{align}x=100, y=2\end{align}

\begin{align}(102)^{3} &\!\!=\!\!(100)^{3}\!\!+\!(2)^{3}\!\!+\!3(100)\!(2)\!(100\!+\!\!2) \\ &=1000000+8+600 \times 102 \\ &=1000008+61200 \\ &=10,61,208 \end{align}

(iii)\begin{align}\;\;(998)^{3}=(1000-2)^{3}\end{align}

Identity: \begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}

Take \begin{align}x=1000, y=-2\end{align}

\begin{align}(998)^{3} &\!=\!(1000)^{3}\!\!\!-\!\!(2)^{3}\!\!\!-\!\!3(1000)\!(2)\!(1000\!-\!\!2\!) \\ &=1000000000-8-6000\,\times\,998 \\ &=999999992+5988000 \\ &=99,40,11,992 \end{align}

## Chapter 2 Ex.2.5 Question 8

Factorise each of the following:

(i) \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align}

(ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align}

(iii) \begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align}

(iv) \begin{align}64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align}

(v) \begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align}

### Solution

Reasoning:

Identities:

\begin{align}&(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y) \\ &{(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)}\end{align}

Steps:

(i) \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align}

This can be re-written as:

\begin{align}(2 a)^{3}+(b)^{3}+3(2 a)^{2}(b)+3(2 a)(b)^{2}\end{align}

Which is of the form:

\begin{align}x^{3}+y^{3}+3 x y(x+y)=(x+y)^{3}\end{align}

Hence,

\begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}=(2 a+b)^{3}\end{align}

(ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align}

This can be re-written as:

\begin{align}(2 a)^{3}-(b)^{3}-3(2 a)^{2}(b)+3(2 a)(b)^{2}\end{align}

Which is of the form:

\begin{align}x^{3}-y^{3}-3 x^{2} y+3 x y^{2}=(x-y)^{3}\end{align}

Hence,

\begin{align}8 a^{3}\!-\!b^{3}\!-\!12 a^{2} b\!+\!6 a b^{2}\!=\!(2 a\!-\!b)^{3}\end{align}

(iii)\begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align}

This can be re-witten as:

\begin{align}&(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2} \\&(3)^{3}-(5 a)^{3}-3(3)(5 a)(3-5 a) & \end{align}

Which is of the form:

\begin{align} x^{3}-y^{3}-3 x y(x-y)=(x-y)^{3}\end{align}

Hence,

\begin{align}27-125 a^{3}-135 a+225 a^{2}=(3-5 a)^{3}\end{align}

(iv) \begin{align} 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align}

This can be re-written as:

\begin{align}&(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2} \\ &(4 a)^{3}-(3 b)^{3}-3(4 a)(3 b)(4 a-3 b) \end{align}

Which is of the form:

\begin{align}x^{3}-y^{3}-3 x y(x-y)=(x-y)^{3}\end{align}

Hence,

\begin{align}64 a^{3}\!-\!27 b^{3}\!-\!144 a^{2} b\!+\!108 a b^{2}\!=\!(4 a\!-\!3 b)^{3}\end{align}

(v) \begin{align}\;\;27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align}

This can be re-written as:

\begin{align}&(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2} \frac{1}{6}+3(3 p)\left(\frac{1}{6}\right)^{2} \\ &(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p) \frac{1}{6}\left(3 p-\frac{1}{6}\right)\end{align}

Which is of the form:

\begin{align}a^{3}-b^{3}-3 a b(a-b)=(a-b)^{3}\end{align}

Hence,

\begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p=\left(3 p-\frac{1}{6}\right)^{3}\end{align}

## Chapter 2 Ex.2.5 Question 9

Verify:

(i) \begin{align}\left(x^{3}+y^{3}\right)=(x+y)\left(x^{2}-x y+y^{2}\right) \end{align}

(ii) \begin{align}\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+x y+y^{2}\right)\end{align}

### Solution

Steps:

(i) \begin{align}\left(x^{3}+y^{3}\right)=(x+y)\left(x^{2}-x y+y^{2}\right) \end{align}

\begin{align}&(x\!+\!y) \left(x^{2}\!-\!x y\!+\!y^{2}\right) \\&= x\left(x^{2}\!-\!x y\!+\!y^{2}\right)\!+\!y\left(x^{2}\!-\!x y\!+\!y^{2}\right) \\&\!=\! x^{3}\!-\!x^{2} y\!+\!x y^{2}\!+\!x^{2} y\!-\!x y^{2}\!+\!y^{3} \\&=\!x^{3}\!+\!y^{3} \end{align}

(ii) \begin{align}\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+x y+y^{2}\right)\end{align}

\begin{align}&{(x\!-\!y)\left(x^{2}\!+\!x y\!+\!y^{2}\right)} \\ &\!=\!x\left(x^{2}\!+\!x y\!+\!y^{2}\right)\!-\!y\left(x^{2}\!+\!x y\!+\!y^{2}\right) \\ &=x^{3}\!+\!x^{2} y\!+\!x y^{2}\!-\!x^{2} y\!-\!1x y^{2}-\!y^{3} \\ &=x^{3}-y^{3} \end{align}

## Chapter 2 Ex.2.5 Question 10

Factorise each of the following:

(i) \begin{align}27 y^{3}+125 z^{3} \end{align}

(ii) \begin{align}64 m^{3}-343 n^{3}=(4 m)^{3}-(7 n)^{3} \end{align}

[Hint: See Question 9.]

### Solution

Steps:

(i) \begin{align}27 y^{3}+125 z^{3}=(3 y)^{3}+(5 z)^{3}\end{align}

Using factorization:

\begin{align}\left(x^{3}+y^{3}\right)=(x+y)\left(x^{2}-x y+y^{2}\right)\end{align}

We can write:

\begin{align}&(3 y)^{3}+(5 z)^{3} \\&=(3 y+5 z)\left[(3 y)^{2}-(3 y)(5 z)+(5 z)^{2}\right]\end{align}

\begin{align}&27 y^{3}+125 z^{3}\\&=(3 y+5 z)\left(9 y^{2}-15 y z+25 z^{2}\right)\end{align}

(ii) \begin{align}64 m^{3}-343 n^{3}=(4 m)^{3}-(7 n)^{3}\end{align}

Using factorization:

\begin{align}\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+x y+y^{2}\right)\end{align}

We can write:

\begin{align}&{(4 m)^{3}-(7 n)^{3}}\\&={(4 m-7 n)\left[(4 m)^{2}+(4 m)(7 n)+(7 n)^{2}\right]} \end{align}

\begin{align}& {64 m^{3}-343 n^{3}}\\&={(4 m-7 n)\left(16 m^{2}+28 m n+49 n^{2}\right)}\end{align}

## Chapter 2 Ex.2.5 Question 11

Factorise: \begin{align}27 x^{3}+y^{3}+z^{3}-9 x y z\end{align}

### Solution

Reasoning:

Identity:

\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&\!=\!(x\!+\!y\!+\!z)\!\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!z x\right)\end{align}

Steps:

The above expression can be written as:

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

By using the identity:

\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&=(x\!+\!y\!+\!z)\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!z x\right)\end{align}

We can write:

\begin{align} &{(3 x)^{3}+(y)^{3}+(z)^{2}-3(3 x)(y)(z)} \\ &=(3 x+y+z)\!\left[\begin{array}((3 x)^{2}+(y)^{2}+(z)^{2}-\3 x)(y)\!-\!y z\!-\!(z)(3 x)\end{array}\right]\end{align} Hence, \begin{align}&27 x^{3}+y^{3}+z^{3}-9 x y z\\&=(3 x+y+z)\left(\begin{array}9 x^{2}+y^{2}+z^{2}-\\3 x y-y z-3 z x\end{array}\right)\end{align} ## Chapter 2 Ex.2.5 Question 12 Verify that: \begin{align}&x^{3}+y^{3}+z^{3}-3 x y\\&=\frac{1}{2}(x+y+z)\left[\begin{array}((x-y)^{2}+(y-z)^{2}\\+(z-x)^{2}\end{array}\right]\end{align} ### Solution Reasoning: Identity: \begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&=(x+y+z)\left(\begin{array}(x^{2}+y^{2}+z^{2}-\\x y-y z-z x\end{array}\right)\end{align} Steps: Taking RHS \begin{align}&=\frac{1}{2}(x+y+z)\left[\begin{array}((x-y)^{2}+(y-z)^{2}\\+(z-x)^{2}\end{array}\right] \\ &=\frac{1}{2}(x+y+z)\left[\begin{array}((x^{2}-2 x y+y^{2})+\\(y^{2}-2 y z+z^{2})+\\(z^{2}-2 z x+x^{2})\end{array}\right]\\ &=\frac{1}{2}(x+y+z)\left[\begin{array}(2 x^{2}+2 y^{2}+2 z^{2}-\\2 x y-2 y z-2 z x\end{array}\right] \\ &=\frac{1}{2}(x+y+z)(2)\left[\begin{array}(x^{2}+y^{2}+z^{2}-\\x y-y z-z x\end{array}\right]\\&=\!\begin{Bmatrix}\!x\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\!+\!\!\!\\\!y\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\!+\!\!\!\\z\left[x^{2}+y^{2}+z^{2}-x y-y z-z x\right]\end{Bmatrix}\\ &=\left[\begin{array}(x^{3}+x y^{2}+x z^{2}-x^{2} y-x y z-\\x^{2} z+x^{2} y +y^{3}+y z^{2}-x y^{2}-\\y^{2} z-x y z+z x^{2}+y^{2} z+z^{3}-\\x y z-y z^{2}-x z^{2} \end{array}\right] \\ &=x^{3}+y^{3}+z^{3}-3 x y z=\mathrm{LHS}\end{align} ## Chapter 2 Ex.2.5 Question 13 If \(\begin{align}x+y+z=0,\end{align}

show that  \begin{align}x^{3}+y^{3}+z^{3}=3 x y z\end{align}

### Solution

Reasoning:

Identity:

\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&=\!(x\!+\!y\!+\!z)\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!zx\right)\end{align}

Steps:

By the identity:

\begin{align}&x^{3}+y^{3}+z^{3}-3 x y z\\&=\!(x\!+\!y\!+\!z)\left(x^{2}\!+\!y^{2}\!+\!z^{2}\!-\!x y\!-\!y z\!-\!zx\right)\end{align}

If  $$x + y + z = 0$$  then the entire $$RHS$$ becomes $$0$$ and hence the $$LHS$$

\begin{align}x^{3}+y^{3}+z^{3}-3 x y z=0\end{align}

Hence,

\begin{align}x^{3}+y^{3}+z^{3}=3 x y z\end{align}

## Chapter 2 Ex.2.5 Question 14

Without actually calculating the cubes, find the value of each of the following:

(i) \begin{align}(-12)^{3}+(7)^{3}+(5)^{3}\end{align}

(ii) \begin{align}(28)^{3}+(-15)^{3}+(-13)^{3}\end{align}

### Solution

Reasoning:

If \begin{align}x+y+z=0\end{align} then

\begin{align}x^{3}+y^{3}+z^{3}=3 x y z\end{align}

Steps:

(i) Let \begin{align}x&=-12, y=7, z=5 \end{align}

Then\begin{align}x+y+z=-12+7+5=0\end{align}

So by using the identity,

\begin{align}(-12)^{3}+(7)^{3}+(5)^{3} &=3(-12)(7)(5) \\ &=-1260 \end{align}

(ii)  Let $$x$$ \begin{align}&=28, y=-15, z=-13 \end{align}

Then \begin{align}x+y+z=28-15-13=0 \end{align}

So by using the identity,

\begin{align}&(28)^{3}+(-15)^{3}+(-13)^{3} \\&=3(28)(-15)(-13) \\ &=16380 \end{align}

## Chapter 2 Ex.2.5 Question 15

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: $$25 a^{2}-35 a+12$$

(ii) Area: $$35 y^{2}+13 y-12$$

### Solution

Reasoning:

Area of rectangle $$=$$ Length $$\times$$ Breadth

What is known?

Area of rectangle

What is  unknown?

Length and breadth of the rectangle.

Steps:

Area of rectangle $$=$$ Length $$\times$$ Breadth

Hence, we shall factorise the given expression

$$25 a^{2}-35 a+12$$

Now taking  $$25 a^{2}-35 a+12$$

find $$2$$ numbers $$p, q$$ such that:

(i) $$p+q=$$  co-efficient of $$a$$

(ii) $$p q=$$  co-efficient of $$a^{2}$$ and the constant term.

$$p+q=-35$$  (co-efficient of a )

$$p q=25 \times 12=300$$  (co-efficient of $$a^{2}$$ and the constant term.)

By trial and error method, we get

$$p=-20, q=-15$$

Now splitting the middle term of the given polynomial,

\begin{align}&25 a^{2}-35 a+12 \\&=25 a^{2}-20 a-15 a+12 \\ &=25 a^{2}-15 a-20 a+12 \\ &=5 a(5 a-3)-4(5 a-3) \\ &=(5 a-4)(5 a-3) \end{align}

\begin{align}\therefore 25 a^{2}-35 a+12 &=(5 a-4)(5 a-3)\end{align}

Length $$=$$ $$5 a-3$$  Breadth$$=5 a-4$$

Length$$=5 a-4$$ Breadth$$=5 a-3$$

What is known?

Area of rectangle.

What is unknown?

Length and breadth of the rectangle.

Steps:

Area of rectangle $$=$$ Length $$\times$$ Breadth

Hence, we shall factorise the given expression:

$$35 y^{2}+13 y-12$$

Now taking $$35 y^{2}-13 y-12$$ ,

find $$2$$ numbers $$p, q$$ such that:

i. $$p+q=$$ co-efficient of $$y$$

ii. $$p q=$$ co-efficient of $$y^{2}$$  and the constant term.

$$p+q= -13$$ (co-efficient of  y )
$${p q}=35 \times-12=-420$$(co-efficient of $$y^{2}$$ and the constant term.)

By trial and error method, we get

$$p = -28, q = -15.$$

Now splitting the middle term of the given polynomial,

\begin{align}& 35 y^{2}+13 y-12\\ &=35 y^{2}+28 y-15 y-12 \\ &=7 y(5 y+4)-3(5 y+4) \\ &=(5 y+4)(7 y-3)\end{align}

\begin{align} \therefore 35 y^{2}+13 y-12 &=(5 y+4)(7 y-3)\end{align}

Length $$=5 y+4$$  Breadth $$=7 y-3$$

Length$$=7 y-3$$ Breadth $$=5 y+4$$

## Chapter 2 Ex.2.5 Question 16

What are the possible expressions for the dimensions of the cuboids whose volume are given below?

(i). Volume:  $$3 x^{2}-12 x$$

(ii). Volume: $$12 k y^{2}+8 k y-20 k$$

### Solution

Reasoning:

(i) Volume of a cubiod = length $$\times$$ breadth $$\times$$ height

What is known?

Volume of cubiod.

What is unknown?

Length, breadth and height of the cuboid.

Steps:

Volume of a cubiod = length $$\times$$ breadth $$\times$$ height

Hence we shall express the given polynomial as product of three expression.

$$3 x^{2}-12 x=3 x(x-4)$$

Length $$= 3$$, breadth $$= x$$, height $$= x-4$$

Length $$= 3$$, breadth $$= x-4$$, height $$= x$$

Length $$= x$$, breadth $$= 3$$, height $$= x-4$$

Length $$= x-4$$, breadth $$= x-4$$, height $$= 3$$

Length $$= x-4$$, breadth $$= 3$$, height $$= x$$

Length $$= x-4$$, breadth $$= 3$$, height$$= x$$

(ii)

What is known?

Volume of cubiod.

What is unknown?

Length, breadth and height of the cuboid.

Steps:

Volume of a cubiod = length $$\times$$ breadth $$\times$$ height

Hence, we shall express the given polynomial as product of three factors

$$12 k y^{2}+8 k y-20 k=4 k\left(3 y^{2}+2 y-5\right)$$

Now taking $$3 y^{2}+2 y-5$$,

find$$2$$  numbers $$p, q$$ such that:

i.  $$p+q=$$co-efficient of  $$y$$

ii .$$p q=$$co-efficient of  $$y^{2}$$ and the constant term.

$$p+q=2$$(co-efficient of y)
$$p q=3 \times-5=-15$$(co-efficient of  $$y^{2}$$ and the constant term.)

By trial and error method, we get

$$p = 5, q = -3.$$

Now splitting the middle term of the given polynomial,

\begin{align} 3 y^{2}+2 y-5 &=3 y^{2}+5 y-3 y-5 \\ &=3 y^{2}-3 y+5 y-5 \\ &=3 y(y-1)+5(y-1) \\ &=(3 y+5)(y-1)\end{align}
Volume $$=4 k(y-1)(3 y+5)$$

Length $$=4 k$$, breadth $$=y-1$$, height $$=3 y+5$$

Length $$=4 k$$, breadth $$=3 y+5$$, height$$=y-1$$

Length $$=y-1$$, breadth $$=4 k$$, height $$=3 y+5$$

Length $$=y-1$$, breadth $$=3y+5 ,$$ height $$=4 k$$

Length $$=3 y+5$$, breadth $$=4 k$$, height $$=y-1$$

Length $$=3 y+5$$, breadth$$=y-1$$ , height $$=4 k$$

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