Ncert Solutions Class 9 Maths Chapter 2 Exercise 2.5

\( (x+4)(x+10) \)
Using identity,
\((x+a)(x+b)=x^{2}+(a+b) x+a b\)
In \( (x+4)(x+10), a=4 \) and \( b=10 \)
Now, \( (x+4)(x+10)=x^{2}+(4+10) x+(4 \times 10) \)
\(=x^{2}+14 x+40\)

\( (\mathrm{x}+8)(\mathrm{x}-10) \)
Using identity,
\((x+a)(x+b)=x^{2}+(a+b) x+a b\)
Here, \( a=8 \) and \( b=-10 \)
\((x+8)(x-10)=x^{2}+\{8+(-10)\} x+\{8 \times(-10)\}\)
\(=x^{2}+(8-10) x-80\)
\(=x^{2}-2 x-80\)

\( (3 x+4)(3 x-5) \)
Using identity,
\((x+a)(x+b)=x^{2}+(a+b) x+a b\)
Here, \( x=3 x, a=4 \) and \( b=-5 \)
\((3 x+4)(3 x-5)=(3 x)^{2}+\{4+(-5)\} 3 x+\{4 \times(-5)\}\)
\(=9 x^{2}+3 x(4-5)-20\)
\(=9 x^{2}-3 x-20\)

\( \left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\)
Using identity,
\((x+y)(x-y)=x^{2}-y^{2}\)
Here, \( x=y^{2} \) and \( y=\frac{3}{2} \)
\(\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\)
\(=y^{4}-\frac{9}{4}\)

\( (3-2 x)(3+2 x) \)
using identity
\((x+y)(x-y)=x^{2}-y^{2}\)
Here, \( x=3 \) and \( y=2 x \)
\((3-2 x)(3+2 x)=3^{2}-(2 x)^{2}\)
\(=9-4 x^{2}\)

Using identity,
\((x+a)(x+b)=x^{2}+(a+b) x+a b\)
Here, \( x=100, a=3 \) and \( b=7 \)
\(103 \times 107=(100+3)(100+7)\)
\(=(100)^{2}+(3+7) 100+(3 \times 7)\)
\(=10000+1000+21\)
\(=11021\)

Using identity,
\((x-a)(x-b)=x^{2}-(a+b) x+a b\)
Here, \( x=100, a=5 \) and \( b=4 \)
\(95 \times 96=(100-5)(100-4)\)
\(=(100)^{2}-(5+4) 90+(5 \times 4)\)
\(=10000-1800+20\)
\(=9120\)

Using identity,
\((x+y)(x-y)=x^{2}-y^{2}\)
Here, \( x=100 \) and \( y=4 \)
\(104 \times 96=(100+4)(100-4)\)
\(=(100)^{2}-(4)^{2}\)
\(=10000-16\)
\(=9984\)

\( 9 x^{2}+6 x y+y^{2} \)
\(=(3 \mathrm{x})^{2}+(2 \times 3 \mathrm{x} \times \mathrm{y})+\mathrm{y}^{2}\)
Using identity,
\((a+b)^{2}=a^{2}+2 a b+b^{2}\)
Here, \( \mathrm{a}=3 \mathrm{x} \) and \( \mathrm{b}=\mathrm{y} \)
\(9 x^{2}+6 x y+y^{2}=(3 x)^{2}+(2 \times 3 \times y)+y^{2}\)
\(=(3 x+y)^{2}\)
\(=(3 x+y)(3 x+y)\)

\( 4 y^{2}-4 y+1=(4 y)^{2}-(2 \times 2 y \times 1)+12 \)
Using identity, \( (a-b)^{2}=a^{2}-2 a b+b^{2} \)
Here, \( \mathrm{a}=2 \mathrm{y} \) and \( \mathrm{b}=1 \)
\(4 y^{2}-4 y+1=(2 y)^{2}-(2 \times 2 y \times 1)+12\)
\(=(2 y-1)^{2}\)
\(=(2 y-1)(2 y-1)\)

Using identity,
\(a^{2}-b^{2}=(a+b)(a-b)\)
Here, \( \mathrm{a}=\mathrm{x} \) and \( \mathrm{b}=\left(\frac{y}{10}\right) \)
\(\mathrm{x}^{2}-\frac{y^{2}}{100}=\mathrm{x}^{2}-\left(\frac{y}{10}\right)^{2}\)
\(=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)\)

Using identity,
\((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( a=x, b=2 y \) and \( c=4 z \)
\((x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+(2 \times x \times 2 y)+(2 \times 2 y \times 4 z)+(2\times 4 z \times x)\)
\(=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z\)

Using identity,
\((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( a=2 x, b=-y \) and \( c=z \)
\((2 x-y+z)^{2}\)
\(=(2 x)^{2}+(-y)^{2}+z^{2}+(2 \times 2 x \times-y)+(2 \times-y \times z)+(2 \times z \times 2x)\)
\(=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z\)

Using identity, \( (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( a=-2 x, b=3 y \) and \( c=2 z \)
\((-2 \mathrm{x}+3 \mathrm{y}+2 \mathrm{z})^{2}\)
\(=(-2 \mathrm{x})^{2}+(3 \mathrm{y})^{2}+(2\mathrm{z})^{2}+(2 \times-2 \mathrm{x} \times 3 \mathrm{y})+(2 \times 3 \mathrm{y} \times 2\mathrm{z}) +(2 \times 2 \mathrm{z} \times-2 \mathrm{x})\)
\(=4 \mathrm{x}^{2}+9 \mathrm{y}^{2}+4 \mathrm{z}^{2}-12 \mathrm{xy}+12 \mathrm{yz}-8\mathrm{xz}\)

Using identity, \( (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( \mathrm{a}=3 \mathrm{a}, \mathrm{b}=-7 \mathrm{b} \) and \( \mathrm{c}=-\mathrm{c}\)
\((3 a-7 b-c)^{2}\)
\(=(3 a)^{2}+(-7 b)^{2}+(-c)^{2}+(2 \times 3 a \times-7 b)+(2 \times-7 b\times-c)+(2 \times-c \times 3 a)\)
\(=9 a^{2}+49 b^{2}+c^{2}-42 a b+14 b c-6 a c\)

Using identity, \( (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( \mathrm{a}=-2 \mathrm{x}, \mathrm{b}=5 \mathrm{y} \) and \( \mathrm{c}=-3 \mathrm{z}\)
\((-2 \mathrm{x}+5 \mathrm{y}-3 \mathrm{z})^{2}\)
\(=(-2 \mathrm{x})^{2}+(5 \mathrm{y})^{2}+(-3\mathrm{z})^{2}+(2 \times-2 \mathrm{x} \times 5 \mathrm{y})+(2 \times 5 \mathrm{y} \times-3\mathrm{z}) +(2 \times-3 \mathrm{z} \times-2 \mathrm{x})\)
\(=4 \mathrm{x}^{2}+25 \mathrm{y}^{2}+9 \mathrm{z}^{2}-20 \mathrm{xy}-30 \mathrm{yz}+12\mathrm{xz}\)

Using identity, \( (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
Here, \( a=\frac{1}{4} \mathrm{a}, \mathrm{b}=\frac{-1}{2} \mathrm{b} \) and \( \mathrm{c}=1 \)
\(\left(\frac{1}{4} a+\frac{-1}{2} b+1\right)^{2}\)
\(=\left(\frac{1}{4}a\right)^{2}+\left(\frac{-1}{2} b\right)^{2}+(1)^{2}+\left(2 \times \frac{1}{4} a \times\frac{-1}{2} b\right)+ \left(2 \times \frac{-1}{2} b \times 1\right)\left(2 \times 1 \times \frac{1}{4}a\right)\)
\(=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a\)

Using identity,
\( (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
\(4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z\)
\(=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+(2 \times 2 \times 3 y)+(2 \times 3 y \times-4 z)+(2 \times-4 z \times 2 x)\)
\(=(2 x+3 y-4 z)^{2}\)
\(=(2 x+3 y-4 z)(2 x+3 y-4 z) \)

Using identity,
\((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\)
\(2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{ } 2 x y+4 \sqrt{2 } y z-8 x z\)
\(=(-\sqrt{ 2} x)^{2}+(y)^{2}+(2 \sqrt{2 } z)^{2}+(2 \times-\sqrt{ 2} x \times y)+(2 \times y \times 2 \sqrt{ 2} z)+(2 \times 2 \sqrt{2 } z \times-\sqrt{ 2} x)\)
\(=(-\sqrt{ 2} x+y+2 \sqrt{2 } z)^{2}\)
\(=(-\sqrt{ 2} x+y+2 \sqrt{ 2} z)(-\sqrt{ 2} x+y+2 \sqrt{2 } z)\)

\( (2 \mathrm{x}+1)^{3} \)
Using identity,
\((a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\)
\((2 x+1)^{3}=(2 x)^{3}+(1)^{3}+(3 \times 2 x \times 1)(2 x+1)\)
\(=8 x^{3}+1+6 x(2 x+1)\)
\(=8 x^{3}+12 x^{2}+6 x+1\)

Using identity,
\((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\)
\((2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-(3 \times 2 a \times 3 b)(2 a-3 b)\)
\(=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)\)
\(=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}\)

Using identity,
\((a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\)
\(\left(\frac{3}{2} \mathrm{x}+1\right)^{3}=(\mathrm{x})^{3}+1^{3}+(3 \times \mathrm{x} \times1)(\mathrm{x}+1)\)
\(=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2}x\)
\(=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1\)

Using identity
\((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\)
\(\left(x-\frac{2}{3} y\right)^{3}=(x)^{3}-\left(\frac{2}{3} y\right)^{3}-\left(3 \times x\times \frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)\)
\(=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)\)
\(=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}\)

\( (99)^{3}=(100-1)^{3} \)
Using identity,
\((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\)
\((100-1)^{3}=(100)^{3}-1^{3}-(3 \times 100 \times 1)(100-1)\)
\(=1000000-1-300(100-1)\)
\(=1000000-1-30000+300\)
\(=970299\)

\( (102)^{3}=(100+2)^{3} \)
Using identity,
\((a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\)
\((100+2)^{3}=(100)^{3}+2^{3}+(3 \times 100 \times 2)(100+2)\)
\(=1000000+8+600(100+2)\)
\(=1000000+8+60000+1200\)
\(=1061208\)

\( (998)^{3}=(1000-2)^{3} \)
Using identity,
\((a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\)
\((1000-2)^{3}=(1000)^{3}-2^{3}-(3 \times 1000 \times 2)(1000-2)\)
\(=1000000000-8-6000(1000-2)\)
\(=1000000000-8-6000000+12000\)
\(=994011992\)

\( 8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2} \)
Using identity:
\((a+b)^{3}=a^{3}+b^{3}+3 a^{2} \mathrm{b}+3 a b^{2}\)
\(=8 a^{3}+b^{3}+12 a^{2} \mathrm{b}+6 a b^{2}\)
\(=(2 a)^{3}+b^{3}+3(2 a)(2 b)+3(2 a)(b)^{2}\)
\(=(2 a+b)^{3}\)
\(=(2 \mathrm{a}+\mathrm{b})(2 \mathrm{a}+\mathrm{b})(2 \mathrm{a}+\mathrm{b})\)

\( 8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2} \)
Using identity,
\((a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}\)
\(=8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\)
\(=(2 a)^{3}-b^{3}-3(2 a)(2 b)+3(2 a)(b)^{2}\)
\(=(2 a-b)^{3}\)
\(=(2 a-b)(2 a-b)(2 a-b)\)

\( 27-125 a^{3}-135 a+225 a^{2} \)
Using identity,
\((a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}\)
\(27-125 a^{3}-135 a+225 a^{2}\)
\(=(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2}\)
\(=(3-5 a)^{3}\)
\(=(3-5 a)(3-5 a)(3-5 a)\)

\( 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2} \)
Using identity
\(64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\)
\(=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}\)
\(=(4 a-3 b)^{3}\)
\(=(4 a-3 b)(4 a-3 b)(4 a-3 b)\)

\( 27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p \)
Using identity
\((a+b)^{3}=a^{3}+b^{3}+3 a^{2} \mathrm{b}+3 a b^{2}\)
\(27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} \mathrm{p}\)
\(=(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2}\left(\frac{1}{6}\right)+3(3p)\left(\frac{1}{6}\right)^{2}\)
\(=\left(3 p-\frac{1}{6}\right)^{3}\)
\(=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\)

\( x^{3}+y^{3}=(x+y)\left(x^{2}+y^{2}-x y\right) \)
we know that
\((x+y)^{3}=(x+y)^{3}-3 x y(x+y)\)
\(\Rightarrow\mathrm{x}^{3}+\mathrm{y}^{3}=(\mathrm{x}+\mathrm{y})\left(\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{xy}\right)\)
[Taking \( (x+y) \) common]
\(\Rightarrow x^{3}+y^{3}=(x+y)\left[(x+y)^{2}-3 x y\right]\)
\(\Rightarrow x^{3}+y^{3}=(x+y)\left[\left(x^{2}+y^{2}+2 x y\right)-3 x y\right] \ldots \ldots\left[(x+y)^{2}=x^{2}+y^{2}+2 x y\right]\)

\( x^{3}-y^{3}=(x-y)\left(x^{2}+y^{2}+x y\right) \)
We know that,
\(\Rightarrow(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\)
Taking \((x-y)\) common
\(\Rightarrow x^{3}-y^{3}=(x-y)\left[(x-y)^{2}+3 x y\right]\)
\(\Rightarrow x^{3}-y^{3}=(\mathrm{x}-\mathrm{y})\left[\left(x^{2}+y^{2}-2 x y\right)+3\mathrm{xy}\right] \ldots \ldots \left[(x-y)^{2}=x^{2}+y^{2}-2 x y\right]\)
\(=x^{3}-y^{3}=(x-y)\left(x^{2}+y^{2}+x y\right)\)

Using identity,
\(x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\)
\(27 y^{3}+125 z^{3}=(3 y)^{3}+(5 z)^{3}\)
\(=(3 y+5 z)\left\{(3 y)^{2}-(3 y)(5 z)+(5 z)^{2}\right\}\)
\(=(3 y+5 z)\left(9 y^{2}-15 y z+25 z^{2}\right)\)

Using identity,
\(x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\)
\(64 m^{3}+125 n^{3}=(4 m)^{3}-(7 n)^{3}\)
\(=(4 m-7 n)\left\{(4 m)^{2}-(4 m)(7 n)+(7 n)^{2}\right\}\)
\(=(4 m-7 n)\left(16 m^{2}+28 m n+49 n\right)^{2}\)

\(27 x^{3}+y^{3}+z^{3}-9 x y z\)
\(=(3 x)^{3}+y^{3}+z^{3}-3 \times 3 x y z\)
Using identity,
\(x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)\)
\(27 x^{3}+y^{3}+z^{3}-9 x y z\)
\(=(3 \mathrm{x}+\mathrm{y}+\mathrm{z})\left\{(3 x)^{2}+y^{2}+z^{2}-3 x y-y z-x z\right\}\)
\(=(3 \mathrm{x}+\mathrm{y}+\mathrm{z})\left(9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z\right)\)

We know that:
\(x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)\)
Dividing and Multiplying RHS by 2 gives,
\(x^{3}+y^{3}+z^{3}-3 x y=\frac{1}{2}(x+y+z) \times 2\left(x^{2}+y^{2}+z^{2}-x y-y z-xz\right)\)
\(=\frac{1}{2}(x+y+z)\left(2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right)\)
\(=\frac{1}{2}(x+y+z)\left[\left(x^{2}+y^{2}-2 x y\right)+\left(y^{2}+z^{2}-2 yz\right)+\left(z^{2}+x^{2}-\right.\right. 2 z x)]\)
As,
\({\left[(x-y)^{2}=x^{2}+y^{2}-2 x y,(y-z)^{2}=y^{2}+z^{2}-2 y z,(z-x)^{2}=z^{2}+x^{2}-2 zx\right]}\)
\(=\frac{1}{2}(x+y+z)\left(2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right)\)
\(=\text { R.H.S. }\)

We know that,
\( x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) \)
Now, put \( (x+y+z)=0 \)
\( x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)\)
\(x^{3}+y^{3}+z^{3}-3 x y z=0\)
\(x^{3}+y^{3}+z^{3}=3 x y z \)
Hence proved.

Let \( \mathrm{x}=-12, \mathrm{y}=7 \) and \( \mathrm{z}=5 \)
We observed that,
\(x+y+z=-12+7+5=0\)
We know that if,
\(x+y+z=0\)
Then,
\(x^{3}+y^{3}+z^{3}=3 x y z\)
\((-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)\)
\(=-1260\)

Let \( \mathrm{x}=28, \mathrm{y}=-15 \) and \( \mathrm{z}=-13 \)
We observed that,
\(x+y+z=28-15-13=0\)
We know that if,
\(x+y+z=0\)
Then,
\(x^{3}+y^{3}+z^{3}=3 x y z\)
\((28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)\)
\(=16380\)

Area: \( 25 \mathrm{a}^{2}-35 \mathrm{a}+12 \)
Since the area is the product of length and breadth therefore by factorizing the given area, we can know the length and breadth of the rectangle
\(25 a^{2}-35 a+12\)
So for factorizing the above equation, we would need two numbers such that, their product is 300 and their sum is 35
\(=25 a^{2}-15 a-20 a+12\)
\(=5 a(5 a-3)-4(5 a-3)\)
\(=(5 a-4)(5 a-3)\)
Possible expression for length \( =(5 \mathrm{a}-4) \) or \( (5 \mathrm{a}-3) \)
Possible expression for breadth \( =(5 a-3) \) or \( (5 \mathrm{a}-4) \)

Area: \( 35 y^{2}+13 y-12 \)
\(35 y^{2}+13 y-12\)
Now we will factorize this equation such that, the product of two numbers is 420 and their difference is 12
\(=35 y^{2}-15 y+28 y-12\)
\(=5 y(7 y-3)+4(7 y-3)\)
\(=(5 y+4)(7 y-3)\)
Possible expression for length \( =(5 y+4) \) or \( (7 y-3) \)
Possible expression for breadth \( =(7 y-3) \) or \( (5 y+4) \)

Volume: \( 3 x^{2}-12 x \)
Since, the volume is the product of length, breadth, and height therefore by factorizing the given volume we can know the length, breadth and height of the cuboid
\(3 x^{2}-12 x\)
\(=3 x(x-4)\)
Possible expression for length \( =3 \)
Possible expression for breadth \( =\mathrm{x} \)
Possible expression for height \( =(\mathrm{x}-4) \)

Volume: \( 12 \mathrm{ky}^{2}+8 \mathrm{ky}-20 \mathrm{k} \)
Since, the volume is the product of length, breadth, and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid
\( 12 \mathrm{ky}^{2}+8 \mathrm{ky}-20 \mathrm{k} \)
\(=4 k\left(3 y^{2}+2 y-5\right)\)
\(=4 k\left(3 y^{2}+5 y-3 y-5\right)\)
\(=4 k[y(3 y+5)-1(3 y+5)]\)
\(=4 k(3 y+5)(y-1)\)
Possible expression for length \( =4 \mathrm{k} \)
Possible expression for breadth \( =(3 y+5) \)
Possible expression for height \( =(\mathrm{y}-1) \)