NCERT Class 9 Maths Exercise 2.4 Solutions

1. Use suitable identities to find the following products:

(x + 4) (x + 10)
Solution:
$$\begin{aligned} \color{blue}\text{Using Identity: } \left( x+a\right) \left( x+b\right) &=\color{blue}x^{2}+\left( a+b\right) x+ab\\\\ a&=4;\\b&=10\\\\ \left( x+4\right) \left( x+10\right) &=x^{2}+\left( 4+10\right) x+10\times 4\\ &=x^{2}+14x+40\end{aligned}$$

$\left( x+8\right) \left( x-10\right)$ $$\begin{aligned} \color{blue}\text{Using Identity: } \left( x+a\right) \left( x+b\right) &=\color{blue}x^{2}+\left( a+b\right) x+ab\\\\ a&=8,\\b&=-10\\\\ \left( x+8\right) \left( x-10\right)&=x^{2}+\left( 8-10\right) x+8\times \left( -10\right) \\ &=x^{2}-2x-80\end{aligned}$$

$\left( 3x+4\right) \left( 3x-5\right)$ $$ \require{cancel}\begin{aligned} \color{blue}\text{Using Identity: }\left( x+a\right) \left( x+b\right)&=\color{blue}x^2+(a+b)x+ab\\ \\\left( 3x+4\right) \left( 3x-5\right) &=3\left( x+\frac{4}{3}\right) \cdot 3\left(x- \frac{5}{3}\right) \\ &=9\left[ \left( x+\frac{4}{3}\right) \left( x-\frac{5}{3}\right) \right] \\\\ &a=\frac{4}{3},\\&b=-\frac{5}{3}\\\\ &=9\left[ x^{2}+\left( \frac{4}{3}-\frac{5}{3}\right) x+\frac{4}{3}\cdot \left( -\frac{5}{3}\right) \right] \\ &=9 \left[ x^{2}+\left\{ \dfrac{\left( 4-5\right) }{3}x\right\} -\dfrac{20}{9}\right] \\ &=9\left[ x^{2}-\dfrac{x}{3}-\dfrac{20}{9}\right] \\ &=9\left[ \dfrac{9x^{2}-3x-20}{9}\right] \\ &=\dfrac{\cancel{9}}{\cancel{9}}\left[9x^{2}-3x-20\right] \\ &=9x^{2}-3x-20\end{aligned}$$

$\left( y^{2}+\dfrac{3}{2}\right) \left( y^{2}-\dfrac{3}{2}\right)$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a+b\right) \left( a-b\right) &=\color{blue}a^{2}-b^{2}\\\\ a&=y^{2},\\b&=\dfrac{3}{2}\\\\ \left( y^{2}+\dfrac{3}{2}\right) \left( y^{2}-\dfrac{3}{2}\right) &=\left( y^{2}\right) ^{2}-\left( \dfrac{3}{2}\right) ^{2}\\ &=y^{4}-\dfrac{9}{4}\end{aligned}$$

$\left( 3-2x\right) \left( 3+2x\right)$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a-b\right) \left( a+b\right) &=\color{blue}a^{2}-b^{2}\\\\ a&=3,\\b&=2x\\\\ \left( 3-2x\right) \left( 3+2x\right) &=3^{2}-\left( 2x\right) ^{2}\\ &=9-4x^{2}\end{aligned}$$

2. Evaluate the following products without multiplying directly:

$103\times 107$ $$\begin{aligned}103\times 107 &=\left( 100+3\right) \left( 100+7\right) \\ \color{blue}\text{Using Identity: }\left( x+a\right) \left( x+b\right) &=\color{blue}x^{2}+\left( a+b\right) x+ab\\\\ x&=100,\\a&=3,\\b&=7\\\\ x^{2}+\left( a+b\right) x+ab &=100^{2}+\left( 3+7\right) 100+3\times 7\\ &=10000+10\times 100+21\\ &=10000+1000+21\\ &=11021\end{aligned}$$

$95\times 96$ $$\begin{aligned}95\times 96 &=\left( 100-5\right) \left( 100-4\right)\\ \color{blue}\text{Using Identity: }\left( x+a\right) \left( x+b\right) &=\color{blue}x^{2}+\left( a+b\right) x+ab\\\\ x&=100,\\a&=-5,\\b&=-4\\\\ x^{2}+\left( a+b\right) x+ab &=\left( 100\right) ^{2}+\left( -5-4\right) 100+\left( -5\right) \left( -4\right) \\ &=10000+\left( -9\right) 100+20\\ &=10000-900+20\\ &=9120\end{aligned}$$

$104\times 96$ $$\begin{aligned}104\times 96 &= \left( 100+4\right) \left( 100-4\right) \\ \color{blue}\text{Using Identity: }\left( x+y\right) \left( x-y\right) &=\color{blue}x^{2}-y^{2}\\\\ x&=100,\\y&=4\\\\ x^{2}-y^{2}\\ =\left( 100\right) ^{2}-\left( 4\right) ^{2}\\ &=10000-16\\ &=9984 \end{aligned}$$

3. Factorise the following using appropriate identities:

$9x^{2}+6xy+y^{2}$ $$\begin{aligned}9x^{2}+6xy+y^{2} &=\left( 3x\right) ^{2}+2\cdot 3xy+y^{2}\\ \color{blue}\text{Using Identity: }a^{2}+2ab+b^{2}&=\color{blue}\left( a+b\right) ^{2}\\\\ a&=3x,\\b&=y\\\\ \left( 3x\right) ^{2}+2\cdot \left( 3x\right) +y^{2} &=\left( 3x+y\right) ^{2}\\ &=\left( 3x+y\right) \left( 3x+y\right) \end{aligned}$$

$4y^{2}-4y+1$ $$\begin{aligned}4y^{2}-4y+1 &=\left( 2y\right) ^{2}-2\cdot \left( 2y \right)\cdot 1 +1^{2}\\ \color{blue}\text{Using Identity: }a^{2}-2ab+b^{2}&=\color{blue}\left( a-b\right) ^{2}\\\\ a&=2y,\\b&=1\\\\ \left( 2y\right) ^{2}-2\left( 2y\right) +1^{2} &=\left( 2y-1\right) ^{2}\\ &=\left( 2y-1\right) \left( 2y-1\right) \end{aligned}$$

$x^{2}-\dfrac{y^{2}}{100}$ $$\begin{aligned}x^{2}-\dfrac{y^{2}}{100}\\ x^{2}-\left( \frac{y}{10}\right) ^{2}\\ \color{blue}\text{Using Identity: }a^{2}-b^{2}&=\color{blue}\left( a+b\right) \left( a-b\right) \\\\ a&=x,\\ b&=y/10\\\\ x^{2}-\left( y/10\right) ^{2}\\ \left( x+\frac{y}{10}\right) \left(x- \frac{y}{10}\right) \end{aligned}$$

4. Expand each of the following, using suitable identities:

$\left( x+2y+4z\right) ^{2}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}&=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\ \left( x+2y+4z\right) ^{2} &=x^{2}+\left( 2y\right) ^{2}+\left( 4z\right) ^{2}+2\cdot x\cdot 2y+2\cdot 2y\cdot 4z+2\cdot 4z\cdot x\\ &=x^{2}+4y^{2}+16z^{2}+4xy+16yz+8zx\end{aligned}$$

$\left( 2x-y+z\right) ^{2}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}&=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\ \left( 2x-y+z\right) ^{2} &=\left( 2x\right) ^{2}+\left( -y\right) ^{2}+z^{2}+2\cdot 2x\cdot \left( -y\right) +2\cdot \left( -y\right) z+2z\cdot 2x\\ &=4x^{2}+y^{2}+z^{2}-4xy-2yz+4zx\end{aligned}$$

$\left( -2x+3y+2z\right) ^{2}\\\\ \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\\\$$ \left( -2x+3y+2z\right) ^{2}\\ \quad =\left( -2x\right) ^{2}+\left( 3y\right) ^{2}+\left( 2z\right) ^{2}+2\cdot \left( -2x\right) \cdot \left( 3y\right) +2\cdot 3y\cdot 2z+2\cdot 2z\cdot \left( -2x\right) \\ \quad =4x^{2}+9y^{2}+4z^{2}-12xy+12yz-8zx$

$ \left( 3a-7b-c\right) ^{2}\\\\ \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\\\$$ \left( 3a-7b-c\right) ^{2}\\ \qquad =\left( 3a\right) ^{2}+\left( -7b\right) ^{2}+\left( -c\right) ^{2}+2\cdot 3a\cdot \left( -7b\right) +2\cdot \left( -7b\right) .\left( -c\right) +2\cdot \left( -c\right) \cdot \left( 3a\right) \\ \qquad =9a^{2}+49b^{2}+c^{2}-42ab+14bc-6ca$

$\left( -2x+5y-3z\right) ^{2}\\\\ \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\\\$$ \left( -2x+5y-3z\right) ^{2}\\ \qquad =\left( -2x\right) ^{2}+\left( 5y\right) ^{2}+\left( -3z\right) ^{2}+2\cdot \left( -2x\right) \cdot \left( 5y\right) +2\cdot \left( 5y\right) \cdot \left( -3z\right) +2\cdot \left( -3z\right) \cdot \left( -2x\right) \\ \qquad =4x^{2}+25y^{2}+9z^{2}-20xy-30yz+12zx$

$\left( \dfrac{1}{4}a-\dfrac{1}{2}b+1\right) ^{2}\\\\ \color{blue}\text{Using Identity: }\left( a+b+c\right) ^{2}=\color{blue}a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\\\\$ $\left( \dfrac{1}{4}a-\dfrac{1}{2}b+1\right) ^{2}\\\\ \qquad =\left( \dfrac{1}{4}a\right) ^{2}+\left( \dfrac{1}{2}b\right) ^{2}+1^{2}+2\cdot \dfrac{1}{4}a\cdot \left( -\dfrac{1}{2}b\right) +2\cdot \left( -\dfrac{1}{2}b\right) \cdot 1+2\cdot 1\cdot \left( \dfrac{1}{4}a\right) \\\\ \qquad =\dfrac{1}{16}a^{2}+\dfrac{1}{4}b^{2}+1+2\cdot \dfrac{1}{4}\cdot \left( -\dfrac{1}{2}\right) ab+2\cdot \left( \dfrac{-1}{2}\right) b+2\cdot \dfrac{1}{4}a\\\\ \qquad =\dfrac{a^{2}}{16}+\dfrac{b^{2}}{4}+1-\dfrac{ab}{4}-b+\dfrac{a}{2}$

5. Factorise:

$4x^{2}+9y^{2}+16z^{2}+12xy+24yz+16xz\\\\$ $4x^{2}+9y^{2}+16z^{2}+12xy+24yz+16xz\\ \qquad =\left( 2x\right) ^{2}+\left( 3y\right) ^{2}+\left( -4z\right) ^{2}+2\cdot 2x\cdot 3y+2\cdot 3y\cdot \left( -4z\right) +2\cdot \left( -4z\right) \left( 2x\right) \\\\ \color{blue}\text{Using Identity: }a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=\color{blue}\left( a+b+c\right) ^{2}\\\\ \qquad =\left( 2x+3y-4z\right) ^{2}\\ \qquad =\left( 2x+3y-4z\right) \left( 2x+3y-4z\right) $

$2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz\\\\$ $2x^{2}+y^{2}+8z^{2}-2\sqrt{2}xy+4\sqrt{2}yz-8xz\\ \qquad =\left( -\sqrt{2}x\right) ^{2}+y^{2}+\left( 2\sqrt{2}z\right) ^{2}-2\cdot \left( \sqrt{2}x\right) \cdot y+2\cdot y\cdot 2\sqrt{2}z-2\cdot \left( 2\sqrt{2}z\right) \cdot \left( -\sqrt{2}x\right) $ $$\color{orange}\left[\because\quad2x^{2}=\left( -\sqrt{2}x\right) ^{2},\quad 8z^{2}=\left( 2\sqrt{2}z\right) ^{2}\right]$$ $\color{blue}\text{Using Identity: }a^{2}+b^{2}+c^{2}+2ab+2bc+2za=\left( a+b+c\right) ^{2}\\\\$ $\qquad =\left( -\sqrt{2}x\right) ^{2}+y^{2}+\left( 2\sqrt{2}z\right) ^{2}-2\cdot \sqrt{2}x\cdot y+2\cdot y\cdot 2\sqrt{2}z-2\cdot \left( 2\sqrt{2}z\right) \cdot \left( -J_{2}x\right) \\ \qquad =\left( -\sqrt{2}x+y+2\sqrt{2}z\right) ^{2}\\ \qquad =\left( -\sqrt{2}x+y+2\sqrt{2}Z\right) \left( -\sqrt{2}x+y+2\sqrt{2}z\right) $

6. Write the following cubes in expanded form:

$\left( 2x+1\right) ^{3}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a+b\right) ^{3}&=\color{blue}a^{3}+b^{3}+3\left( a+b\right) ab\\ \left( 2x+1\right) ^{3} &=\left( 2x\right) ^{3}+1^{3}+3\left( 2x+1\right) \left( 2x\right) \left( 1\right) \\ &=8x^{3}+1+6x(2x+1)\\ &=8x^{3}+1+12x^2 + 6x\\ &=8x^{3}+12x^{2}+6x+1\end{aligned}$$

$\left( 2a-3b\right) ^{3}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a-b\right) ^{3}&=\color{blue}a^{3}-b^{3}-3ab\left( a-b\right) \\\\ \left( 2a-3b\right) ^{3} &=\left( 2a\right) ^{3}-\left( 3b\right) ^{3}-3\left( 2a\right) \left( 3b\right) \left( 2a-3b\right) \\ &=8a^{3}-27b^{3}-18ab\left( 2a-3b\right) \\ &=8a^{3}-27b^{3}-36a^{2}b+54ab^{2}\end{aligned}$$

$\left( \dfrac{3}{2}x+1\right) ^{3}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a+b\right) ^{3}&=\color{blue}a^{3}+b^{3}+3ab\left( a+b\right) \\\\ \left( \dfrac{3}{2}x+1\right) ^{3}&= \left( \dfrac{3}{2}x+1\right) ^{3} \left( \dfrac{3}{2}x\right) ^{3}+1^{3}+3\cdot \dfrac{3}{2}x\cdot 1\left( \dfrac{3}{2}x+1\right) \\\\ &=\dfrac{27}{8}x^{3}+1+\dfrac{9}{2}x\left( \dfrac{3}{2}x+1\right) \\\\ &=\dfrac{27}{8}x^{3}+1+\dfrac{9}{2}x\cdot \dfrac{3}{2}x+\dfrac{9}{2}x\cdot 1\\\\ &=\dfrac{27}{8}x^{3}+1+\dfrac{27x^{2}}{4}+\dfrac{9}{2}x\\\\ &=\dfrac{27}{8}x^{3}+\dfrac{27}{4}x^{2}+\dfrac{9}{2}x+1\end{aligned}$$

$\left( x-\dfrac{2}{3}y\right) ^{3}$ $$\begin{aligned} \color{blue}\text{Using Identity: }\left( a-b\right) ^{3}&=\color{blue}a^{3}-b^{3}-3ab\left( a-b\right) \\\\ \left( x-\dfrac{2}{3}y\right) ^{3} &=x^{3}-\left( \dfrac{2}{3}y\right) ^{3}-3\cdot x\cdot \dfrac{2}{3}y\left( x-\dfrac{2}{3}y\right) \\\\ &=x^{3}-\dfrac{8}{27}y^{3}-2xy\left( x-\dfrac{2}{3}y\right) \\\\ &=x^{3}-\dfrac{8}{27}y^{3}-2x^{2}y+2\cdot \dfrac{2}{3}xy^{2}\\\\ &=x^{3}-\dfrac{8}{27}y^{3}-2x^{2}y+\dfrac{4}{3}xy^{2}\end{aligned}$$

7. Evaluate the following using suitable identities:

$\left( 99\right) ^{3}$ $$\begin{aligned}\left( 99\right) ^{3} &=\left( 100-1\right) ^{3}\\\\ \color{blue}\text{Using Identity: }\left( a-b\right) ^{3}&=\color{blue}a^{3}-b^{3}-3a^2b+3ab^2 \\\\ \left( 100-1\right) ^{3} &=100^{3}-1^{3}-3\cdot 100^2\cdot1+3\cdot100\cdot1^2 \\ &=1000000-1-30000+300 \\ &=1000300-30001\\ &=970299\end{aligned}$$

$\left( 102\right) ^{3}$ $$\begin{aligned}\left( 102\right) ^{3}&=\left( 100+2\right) ^{3}\\\\ \color{blue}\text{Using Identity: }\left( a+b\right) ^{3}&=\color{blue}a^{3}+b^{3}+3ab\left( a+b\right) \\\\ \left( 100+2\right) ^{3} &=\left( 100\right) ^{3}+\left( 2\right) ^{3}+3\cdot 100\cdot 2\left( 100+2\right) \\ &=1000000+8+600\left( 102\right) \\ &=1000000+8+61200\\ &=1000000+61208\\ &=1061208\end{aligned}$$

$\left( 998\right) ^{3}$ $$\begin{aligned}\left( 998\right) ^{3}&=\left( 1000-2\right) ^{3}\\ \color{blue}\text{Using Identity: }\left( a^{3}-b\right) ^{3}&=\color{blue}9^{3}-b^{3}-3a^2b+3ab^2 \\ \left( 1000-2\right) ^{3} &=\left( 1000\right) ^{3}-2^{3}-3\cdot1000^2\cdot 2+3\cdot 1000\cdot 2^2 \\ &=1000000000-8-3\cdot 2\cdot 1000000 + 3\cdot4\cdot1000\\ &=1000000000-8-6000000+12000\\ &=1000012000-6000008\\ &=994011992\end{aligned}$$

8. Factorise each of the following:

$8a^{3}+b^{3}+72a^{2}b+6ab^{2}$ $$\begin{aligned} \color{blue}\text{Using Identity: }a^{3}+b^{3}+3a^{2}b+3ab^{2}&=\color{blue}\left( a+b\right) ^{3}\\\\ 8a^{3}+b^{3}+72a^{2}b+6ab^{2}&=\left( 2a\right) ^{3}+b^{3}+3\cdot \left( 2\right) ^{2}\cdot b+3\left( 2a\right) b^{2}\\ &=\left( 2a+b\right) ^{3}\\ &=\left( 2a+b\right) \left( 2a+b\right) \left( 2a+b\right) \end{aligned}$$

$8a^{3}-b^{3}-12a^{2}b+6ab^{2}$ $$\begin{aligned} \color{blue}\text{Using Identity: }a^{3}-b^{3}-3a^{2}b+3ab^{2}&=\color{blue}\left( a-b\right) ^{3}\\\\ 8a^{3}-b^{3}-12a^{2}b+6ab^{2} &=\left( 2a\right) ^{3}-b^{3}-3\cdot \left( 2a\right) ^{2}b+3\cdot \left( 2a\right) b^{2}\\ &=\left( 2a-b\right) ^{3}\\ &=\left( 2a-b\right) \left( 2a-b\right) \left( 2a-b\right) \end{aligned}$$

$27-125a^{3}-135a+225a^{2}$ $$\begin{aligned} \color{blue}\text{Using Identity: }a^{3}-b^{3}-3a^{2}b+3ab^{2}&=\color{blue}\left( a-b\right) ^{3}\\\\ 27-125a^{3}-135a+225a^{2} &=3^{3}-\left( 5a\right) ^{3}-3\cdot 3^{2}\cdot 5a+3\cdot 3\cdot \left( 5a\right) ^{2}\\ &=\left( 3-5a\right) ^{3}\\ &=\left( 3-5a\right) \left( 3-5a\right) \left( 3-5a\right) \end{aligned}$$

$64a^{3}-27b^{3}-144a^{2}b+108ab^{2}$ $$\begin{aligned}\\ \color{blue}\text{Using Identity: }a^{3}-b^{3}-3a^{2}b+3ab^{2}&=\color{blue}\left( a-b\right) ^{3}\\ 64a^{3}-27b^{3}-144a^{2}b+108ab^{2} &=\left( 4a\right) ^{3}-\left( 3b\right) ^{3}-3\cdot \left( 4a\right) ^{2}\cdot 3b+3\cdot \left( 4a\right) \cdot \left( 3b\right) ^{2}\\ &=\left( 4a-3b\right) ^{3}\\ &=\left( 4a-3b\right) \left( 4a-3b\right) \left( 4a-3b\right) \end{aligned}$$

$27p^{3}-\dfrac{1}{216}-\dfrac{9}{2}p^{2}+\dfrac{1}{4}p\\$ $$\begin{aligned} \color{blue}\text{Using Identity: }a^{3}-b^{3}-3a^{2}b+3ab^{2}&=\color{blue}\left( a-b\right) ^{3}\\\\ 27p^{3}-\dfrac{1}{216}-\dfrac{9}{2}p^{2}+\dfrac{1}{4}p &=\left( 3p\right) ^{3}-\left( \dfrac{1}{6}\right) ^{3}-3-\left( 3p\right) ^{2}\cdot \dfrac{1}{6}+3\cdot 3p\cdot \left( \dfrac{1}{6}\right) ^{2}\\\\ &=\left( 3p-\frac{1}{6}\right) ^{3}\\\\ &=\left( 3p-\frac{1}{6}\right) \left( 3p-\frac{1}{6}\right) \left( 3p-\frac{1}{6}\right) \end{aligned}$$

9. Verify :

$x^{3}+y^{3}=\left( x+y\right) \left( x^{2}-xy+y^{2}\right)\\\\ RHS$ $$\begin{aligned}\left( x+y\right) \left( x^{2}-xy+y^{2}\right) &=x^{3}-x^{2}y+xy^{2}+yx^{2}-xy^{2}+y^{3}\\ &=x^{3}+y^{3}-x^{2}y+x^{2}y+xy^{2}-xy^{2}\\ &=x^{3}+y^{3}+(x^{2}y-x^{2}y)+(xy^{2}-xy^{2})\\ &=x^{3}+y^{3}+(0)+(0)\\ &=x^{3}+y^{3}\end{aligned}$$ $LHS=RHS\quad hence,~verified.$

$x^{3}-y^{3}=\left( x-y\right) \left( x^{2}+xy+y^{2}\right)\\\\RHS$ $$\begin{aligned} \left( x-y\right) \left( x^{2}+xy+y^{2}\right) &=x^{3}+x^{2}y+xy^{2}-yx^{2}-xy^{2}-y^{3}\\ &=x^{3}-y^{3}+x^{2}y-yx^{2}+xy^{2}-xy^{2}\\ &=x^{3}-y^{3}+(x^{2}y-x^{2}y)+(xy^{2}-xy^{2})\\ &=x^{3}-y^{3}+(0)+(0)\\ &=x^{3}-y^{3}\\ RHS=LHS\quad hence,~verified\end{aligned}$$

10. Factorise each of the following:

$27y^{3}+125z^{3}$ $$\begin{aligned}27y^{3}+125z^{3} &=\left( 3y\right) ^{3}+\left( 5z\right) ^{3}\\ \color{blue}\text{Using Identity: }a^{3}+b^{3}&=\color{blue}\left( a+b\right) \left( a^{2}-ab+b^{2}\right) \\\\ \left( 3y\right) ^{3}+\left( 5z\right) ^{3} &=\left( 3y+5z\right) \left\{ \left( 3y\right) ^{2}-3y\cdot 5z+\left( 5z\right) ^{2}\right\} \\ &=\left( 3y+5z\right) \left( 9y^{2}-15yz+25z^{2}\right) \end{aligned}$$

$64m^{3}-343n^{3}$ $$\begin{aligned}64m^{3}-343n^{3}\\ \color{blue}\text{Using Identity: }a^{3}-b^{3}&=\color{blue}\left( a-b\right) \left( a^{2}+ab+b^{2}\right) \\\\ \left( 4m\right) ^{3}-\left( 7n\right) ^{3} &=\left( 4m-7n\right) \left\{ \left( 4m\right) ^{2}+4m\cdot 7n+\left( 7n\right) ^{2}\right\} \\ &=\left( 4m-7n\right) \left\{ 16m^{2}+28mn+49n^{2}\right\} \\ &=\left( 4m-7n\right) \left( 16m^{2}+28mn+49n^{2}\right) \end{aligned}$$

11. Factorise : $27x^3 + y^3 + z^3 – 9xyz$

$$\begin{aligned} \color{blue}\text{Using Identity: }x^{3}+y^{3}+z^{3}-3xyz &=\color{blue}\dfrac{1}{2}\left( x+y+z\right) \left\{ \left( x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-x\right) ^{2}\right\} \\\\ 27x^{3}+y^{3}+z^{3}-9xyz &=\left( 3x\right) ^{3}+y^{3}+z^{3}-3\cdot \left( 3x\right) \cdot y\cdot z\\ &=\dfrac{1}{2}\left( 3x+y+z\right) \left\{ \left( 3x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-3x\right) ^{2}\right\} \end{aligned}$$

12. Verify that $x^{3}+y^{3}+z^{3}-3xyz=\dfrac{1}{2}\left( x+y+z\right) \left[ \left( x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-x\right) ^{2}\right] $

\(RHS:\\\\ \dfrac{1}{2}\left( x+y+z\right) \left[ \left( x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-x\right) ^{2}\right] \\\\ \quad =\dfrac{1}{2}\left( x+y+z\right) \left[ x^{2}+y^{2}-2xy+y^{2}+z^{2}-2yz+z^{2}+x^{2}-2zx\right] \\\\ \quad =\dfrac{1}{2}\left( x+y+z\right) \left[ x^{2}+y^{2}+y^{2}+z^{2}+z^{2}+x^{2}-2xy-2yz-2zx\right] \\\\ \quad =\dfrac{1}{2}\left( x+y+z\right) \left[ 2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx\right] \\\\ \quad =\dfrac{1}{2}\left( x+y+z\right) .2\left[ x^{2}+y^{2}+z^{2}-xy-yz-zx\right] \\\\ \quad =\dfrac{1}{2}\cdot 2\left( x+y+z\right) \left( x^{2}+y^{2}+z^{2}-xy-yz-zx\right) \\\\ \left( x+y+z\right) \left( x^{2}+y^{2}+z^{2}-xy-yz-zx\right) \\\\ \quad =x^{3}+xy^{2}+xz^{2}-x^{2}y-xyz-zx^{2}\\ \qquad\quad+yx^{2}+y^{3}+yz^{2}-xy^{2}-y^{2}z-yzx\\ \qquad\quad+zx^{2}+zy^{2}+z^{3}-zxy-yz^{2}-z^{2}x\\ \quad=x^{3}+y^{3}+z^{3}-xyz-yzx-zxy\\ \quad\quad\quad+(xy^{2}-xy^{2})+(xz^{2}-z^{2}x)+(yx^{2}-x^{2}y)\\ \quad\quad\quad+(zx^{2}-zx^{2})+(yz^{2}-yz^{2})+(zy^{2}-y^{2}z)\\\\ \quad=x^{3}+y^{3}+z^{3}-3xyz\\ \quad\quad\quad+(0)+(0)+(0)\\ \quad\quad\quad+(0)+(0)+(0)\\\\ \quad =x^{3}+y^{3}+z^{3}-3xyz\\ =LHS\\\ \)

13. If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.

$$\begin{aligned}x^{3}+y^{3}+z^{3}-3xyz&=\dfrac{1}{2}\left( x+y+z\right) \left[ \left( x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-x\right) ^{2}\right] \\\\ but~x+y+z&=0\quad \text{as given}\\\\ x^{3}+y^{3}+z^{3}-3xyz&=\dfrac{1}{2}\left( 0\right) \left[ \left( x-y\right) ^{2}+\left( y-z\right) ^{2}+\left( z-x\right) ^{2}\right] \\\\ x^{3}+y^{3}+z^{3}-3xyz&=0\\\\ \Rightarrow x^{3}+y^{3}+z^{3}&=3xyz\end{aligned}$$

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

$\left( -12\right) ^{3}+\left( 7\right) ^{3}+\left( 5\right) ^{3}$ $$\begin{aligned}\left( -12\right) ^{3}+\left( 7\right) ^{3}+\left( 5\right) ^{3}\\ \because\quad-12+7+5&=0\\ \therefore\quad x^{3}+y^{3}+z^{3}&=3xyz\quad\color{blue}\leftarrow\text{Conditional Formula when } x+y+z=0\\ \left( -12\right) ^{3}+\left( 7\right) ^{3}+\left( 5\right) ^{3}&=3\cdot \left( -12\right) \cdot 7\cdot 5\\ &=-60\times 2 \\ &=-1260\end{aligned}$$

(\\left( 28\right) ^{3}+\left( -15\right) ^{3}+\left( -13\right) ^{3}\) $$ \begin{aligned}28-15-13&=0\\ \quad\color{blue}\text{if } (x+y +z&= \color{blue}0)\\\quad\text{then }\Rightarrow x^{3}+y^{3}+z^{3}&=3xyz\\ \therefore \left( 28\right) ^{3}+\left( -15\right) ^{3}+\left( -13\right) ^{3} &=3.28\cdot \left( -15\right) \cdot \left( -13\right)\\ &=3\cdot 28\cdot 15\cdot 13\\ &=16380\end{aligned} $$

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

Area=$A=25a^{2}-35a+12$ $$A=25a^{2}-35a+12$$ $\text{if we can find two numbers } p~and~q \text{ such that } pq=25\times 12~and~ p+q=35,\text{ then by the mid-term split method we can factorise the expression}$ $$\begin{aligned}pq&=12\times 25\\&=3\times 4\times 5\times 5\\&=3\times 5\times 4\times 5\\&=15\times 20\\p&=15,\\q&=20\\ p+q&=15+20\\&=35\\ A&=25a^{2}-35a+12\\&=25a^{2}-15a-20a+12\\&=5a\left( 5a-3\right) -4\left( 5a-3\right) \\&=\left( 5a-3\right) \left( 5a-4\right) =0 \text{ (to find solution we must equate to zero)}\\\Rightarrow5a-3&=0\\a&=\frac{3}{5}\\\Rightarrow 5a-4&=0\\5a&=4 \\a&=\frac{4}{5}\\\text{(Length, Breadth)}&=(3/5, 4/5)\end{aligned}$$

Area=$35y^{2}+13y-12$ $$A=35y^{2}+13y-12$$ $\text{if we can find two numbers } p~and~q \text{ such that } pq=35\times 12~and~ p+q=13,\text{ then by mid-term split method we can factorise the expression}$ $$\begin{aligned}pq &=35\times 12\\ &=5\times 7\times 3\times 4\\ &=15\times 28\\ 35y^{2}+28y-15y-12&=0\\7y\left( 5y+4\right) -3\left( 5y+4\right) &=0\\\left( 5y+4\right) \left( 7y-3\right) &=0\\7y-3&=0\\y&=3/7\\5y+4&=0\\y&=-4/5\end{aligned}$$ $\text{Length cannot be less than zero. }\\ \therefore3/7\text{ is the Answer}$

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

Volume $V=3x^{2}-12x$ $$\begin{aligned}V&=3x^{2}-12x\\ &=3x\left( x-4\right) \\ &=3x\left( x-4\right) \\ &3x\left( x-4\right) =0\\ &3x=0\\ &x-4=0\\ &x=4\end{aligned}$$

Volume $12ky^{2}+8ky-2k$ $$\begin{aligned}12ky^{2}+8ky-2k&=0\\ 12yk^{2}+8yk-20k&=0\\ 4k\left( 3y^{2}+2y-5\right) &=0\\ 3y^{2}+2y-5&=0\\ 3y^{2}-3y+5y-5&=0\\ 3y\left( y-1\right) +5\left( y-1\right) &=0\\ \left( y-1\right) \left( 3y+5\right) &=0\\ y-1&=0\\ y&=1\end{aligned}$$