# RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.2 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 5 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.2 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 5 Factorisation of Algebraic Expressions Exercise 5.2 below and it will be beneficial for them.

## RD Sharma Class 9 Factorisation of Algebraic Expressions Exercise 5.2 Solutions

Q1. $\mathrm{p}^{3}+27$
SOLUTION:
$=\mathrm{p}^{3}+3^{3}$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(p+3)\left(p^{2}-3 p-9\right)$
$\therefore p^{3}+27=(p+3)\left(p^{2}-3 p-9\right)$

Q2. $\mathrm{y}^{3}+125$
SOLUTION :
$=\mathrm{y}^{3}+5^{3}$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(\mathrm{y}+5)\left(\mathrm{y}^{2}-5 \mathrm{y}+5^{2}\right)$
$=(\mathrm{y}+5)\left(\mathrm{y}^{2}-5 \mathrm{y}+25\right)$
$\therefore \mathrm{y}^{3}+125=(\mathrm{y}+5)\left(\mathrm{y}^{2}-5 \mathrm{y}+25\right)$

Q 3 . $1-27 a^{3}$
SOLUTION :
$=(1)^{3}-(3 a)^{3}$
$=(1-3 a)\left(1^{2}+1 \times 3 a+(3 a)^{2}\right)$ $\because\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(1-3 a)\left(1^{2}+3 a+9 a^{2}\right)$
$\therefore 1-27 a^{3}=(1-3 a)\left(1^{2}+3 a+9 a^{2}\right)$

Q 4. $8 x^{3} y^{3}+27 a^{3}$
SOLUTION:
$=(2 x y)^{3}+(3 a)^{3}$
$=(2 x y+3 a)\left((2 x y)^{2}-2 x y \times 3 a+(3 a)^{2}\right)$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(2 x y+3 a)\left(4 x^{2} y^{2}-6 x y a+9 a^{2}\right)$
$\therefore 8 x^{3} y^{3}+27 a^{3}=(2 x y+3 a)\left(4 x^{2} y^{2}-6 x y a+9 a^{2}\right)$

Q5. $64 a^{3}-b^{3}$
SOLUTION:
$=(4 a)^{3}-b^{3}$
$=(4 a-b)\left((4 a)^{2}+4 a \times b+b^{2}\right)$ $\because\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(4 a-b)\left(16 a^{2}+4 a b+b^{2}\right)$
$\therefore 64 a^{3}-b^{3}=(4 a-b)\left(16 a^{2}+4 a b+b^{2}\right)$

Q6 $\cdot \frac{x^{3}}{216}-8 y^{3}$
SOLUTION :
$=\frac{x^{3}}{6}-(2 y)^{3}$
$=\left(\frac{x}{6}-2 y\right)\left(\left(\frac{x}{6}\right)^{2}+\frac{x}{6} \times 2 y+(2 y)^{2}\right)$ $\because\left[x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$
$=\left(\frac{x}{6}-2 y\right)\left(\frac{x^{2}}{36}+\frac{x y}{3}+4 y^{2}\right)$
$\therefore \frac{x^{3}}{216}-8 y^{3}=\left(\frac{x}{6}-2 y\right)\left(\frac{x^{2}}{36}+\frac{x y}{3}+4 y^{2}\right)$

Q7. $10 x^{4} y-10 x y^{4}$
SOLUTION :
$=10 x y\left(x^{3}-y^{3}\right)$
$=10 x y(x-y)\left(x^{2}+x y+y^{2}\right)$ $\because\left[x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$
$\therefore 10 x^{4} y-10 x y^{4}=10 x y(x-y)\left(x^{2}+x y+y^{2}\right)$

Q 8 . $54 x^{6} y+2 x^{3} y^{4}$
SOLUTION :
$=2 x^{3} y\left(27 x^{3}+y^{3}\right)$
$=2 x^{3} y\left((3 x)^{3}+y^{3}\right)$
$=2 x^{3} y(3 x+y)\left((3 x)^{2}-3 x \times y+y^{2}\right)$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=2 x^{3} y(3 x+y)\left(9 x^{2}-3 x y+y^{2}\right)$
$\therefore 54 x^{6} y+2 x^{3} y^{4}=2 x^{3} y(3 x+y)\left(9 x^{2}-3 x y+y^{2}\right)$

Q9. $32 a^{3}+108 b^{3}$
SOLUTION :
$=4\left(8 a^{3}+27 b^{3}\right)$
$=4\left((2 a)^{3}+(3 b)^{3}\right)$
$=4\left[(2 a+3 b)\left((2 a)^{2}-2 a \times 3 b+(3 b)^{2}\right)\right]$
$\because\left[\mathrm{a}^{3}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)\right]$
$=4(2 a+3 b)\left(4 a^{2}-6 a b+9 b^{2}\right)$
$\therefore 32 \mathrm{a}^{3}+108 b^{3}=4(2 a+3 b)\left(4 a^{2}-6 a b+9 b^{2}\right)$

Q $10 \cdot(a-2 b)^{3}-512 b^{3}$
SOLUTION :
$=(a-2 b)^{3}-(8 b)^{3}$
$=(a-2 b-8 b)\left((a-2 b)^{2}+(a-2 b) 8 b+(8 b)^{2}\right)$
$\because\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(a-10 b)\left(a^{2}+4 b^{2}-4 a b+8 a b-16 b^{2}+64 b^{2}\right)$
$=(a-10 b)\left(a^{2}+52 b^{2}+4 a b\right)$
$\therefore(a-2 b)^{3}-512 b^{3}=(a-10 b)\left(a^{2}+52 b^{2}+4 a b\right)$

Q 11 . $(a+b)^{3}-8(a-b)^{3}$
SOLUTION:
$=(a+b)^{3}-[2(a-b)]^{3}$
$=(a+b)^{3}-[2 a-2 b]^{3}$
$=(a+b-(2 a-2 b))\left((a+b)^{2}+(a+b)(2 a-2 b)+(2 a-2 b)^{2}\right)$
$\because\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(a+b-2 a+2 b)\left(a^{2}+b^{2}+2 a b+(a+b)(2 a-2 b)+(2 a-2 b)^{2}\right)$
$=(a+b-2 a+2 b)\left(a^{2}+b^{2}+2 a b+2 a^{2}-2 a b+2 a b-2 b^{2}+(2 a-2 b)^{2}\right)$
$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+(2 a-2 b)^{2}\right)$
$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+4 a^{2}+4 b^{2}-8 a b\right)$
$=(3 b-a)\left(3 a^{2}+4 a^{2}-b^{2}+4 b^{2}-8 a b+2 a b\right)$
$=(3 b-a)\left(7 a^{2}+3 b^{2}-6 a b\right)$
$\therefore(a+b)^{3}-8(a-b)^{3}=(3 b-a)\left(7 a^{2}+3 b^{2}-6 a b\right)$

Q12. $(x+2)^{3}+(x-2)^{3}$
SOLUTION :
$=(x+2+x-2)\left((x+2)^{2}-(x+2)(x-2)+(x-2)^{2}\right)$
$\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=2 x\left(x^{2}+4 x+4-(x+2)(x-2)+x^{2}-4 x+4\right)$
$=2 x\left(2 x^{2}+8-\left(x^{2}-2^{2}\right)\right)$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$=2 x\left(2 x^{2}+8-x^{2}+4\right)$
$=2 x\left(x^{2}+12\right)$
$\therefore(\mathrm{x}+2)^{3}+(\mathrm{x}-2)^{3}=2 \mathrm{x}\left(\mathrm{x}^{2}+12\right)$

Q13. $8 x^{2} y^{3}-x^{5}$
SOLUTION :
$=x^{2}\left((2 y)^{3}-x^{3}\right)$
$=x^{2}(2 y-x)\left((2 y)^{2}+2 y \times x+x^{2}\right)$ $\left[\because x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$
$=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$
$\therefore 8 x^{2} y^{3}-x^{5}=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$

Q 14 . $1029-3 x^{3}$
SOLUTION :
$=3\left(343-x^{3}\right)$
$=3\left((7)^{3}-x^{3}\right)$
$=3(7-x)\left(7^{2}+7 x+x^{2}\right)$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=3(7-x)\left(49+7 x+x^{2}\right)$
$\therefore 1029-3 x^{3}=3(7-x)\left(49+7 x+x^{2}\right)$

Q15. $x^{6}+y^{6}$
SOLUTION :
$=\left(x^{2}\right)^{3}+\left(y^{2}\right)^{3}$
$=\left(x^{2}+y^{2}\right)\left(\left(x^{2}\right)^{2}-x^{2} y^{2}+\left(y^{2}\right)^{2}\right)$
$=\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)$ $\left[\because \mathrm{a}^{3}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)\right]$
$\therefore x^{6}+y^{6}=\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)$

Q16. $x^{3} y^{3}+1$
SOLUTION :
$=(x y)^{3}+1^{3}$
$=(\mathrm{xy}+1)\left((\mathrm{xy})^{2}+\mathrm{xy}+1^{2}\right)$ $\left[\because x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\right]$
$=(x y+1)\left(x^{2} y^{2}-x y+1\right)$
$\therefore x^{3} y^{3}+1=(x y+1)\left(x^{2} y^{2}-x y+1\right)$

Q17. $x^{4} y^{4}-x y$
SOLUTION :
$=x y\left(x^{3} y^{3}-1\right)$
$=x y\left((x y)^{3}-1^{3}\right)$
$=x y(x y-1)\left((x y)^{2}+x y \times 1+1^{2}\right)$ $\because\left[x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$
$=x y(x y-1)\left(x^{2} y^{2}+x y+1\right)$
$\therefore x^{4} y^{4}-x y=x y(x y-1)\left(x^{2} y^{2}+x y+1\right)$

Q 18 . $a^{12}+b^{12}$
SOLUTION :
$=\left(a^{4}\right)^{3}+\left(b^{4}\right)^{3}$
$=\left(a^{4}+b^{4}\right)\left(\left(a^{4}\right)^{2}-a^{4} \times b^{4}+\left(b^{4}\right)^{2}\right)$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=\left(a^{4}+b^{4}\right)\left(a^{8}-a^{4} b^{4}+b^{8}\right)$
$\therefore a^{12}+b^{12}=\left(a^{4}+b^{4}\right)\left(a^{8}-a^{4} b^{4}+b^{8}\right)$

Q 19 . $x^{3}+6 x^{2}+12 x+16$
SOLUTION :
$=x^{3}+6 x^{2}+12 x+8+8$
$=x^{3}+3 \times x^{2} \times 2+3 \times x \times 2^{2}+2^{3}+8$
$=(x+2)^{3}+8$ $\left[\because \mathrm{a}^{3}+3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})^{3}\right]$
$=(x+2)^{3}+2^{3}$
$=(x+2+2)\left((x+2)^{2}-2(x+2)+2^{2}\right)$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(x+2+2)\left(x^{2}+4+4 x-2 x-4+4\right)$ $\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$=(x+4)\left(x^{2}+4+2 x\right)$
$\therefore \mathrm{x}^{3}+6 \mathrm{x}^{2}+12 \mathrm{x}+16=(\mathrm{x}+4)\left(\mathrm{x}^{2}+4+2 \mathrm{x}\right)$

Q 20 . $a^{3}+b^{3}+a+b$
SOLUTION :
$=\left(a^{3}+b^{3}\right)+1(a+b)$
$=(a+b)\left(a^{2}-a b+b^{2}\right)+1(a+b)$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(a+b)\left(a^{2}-a b+b^{2}+1\right)$
$\therefore \mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{a}+\mathrm{b}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}+1\right)$

Q $21 \cdot a^{3}-\frac{1}{a^{3}}-2 a+\frac{2}{a}$
SOLUTION :
$=\left(a^{3}-\frac{1}{a^{3}}\right)-2\left(a-\frac{1}{a}\right)$
$=\left(a^{3}-\left(\frac{1}{a}\right)^{3}\right)-2\left(a-\frac{1}{a}\right)$
$=\left(a-\frac{1}{a}\right)\left(a^{2}+a \times \frac{1}{a}+\left(\frac{1}{a}\right)^{2}\right)-2\left(a-\frac{1}{a}\right)$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=\left(a-\frac{1}{a}\right)\left(a^{2}+1+\frac{1}{a^{2}}\right)-2\left(a-\frac{1}{a}\right)$
$=\left(a-\frac{1}{a}\right)\left(a^{2}+1+\frac{1}{a^{2}}-2\right)$
$=\left(a-\frac{1}{a}\right)\left(a^{2}+\frac{1}{a^{2}}-1\right)$
$\therefore \mathrm{a}^{3}-\frac{1}{\mathrm{a}^{3}}-2 \mathrm{a}+\frac{2}{\mathrm{a}}=\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)\left(\mathrm{a}^{2}+\frac{1}{\mathrm{a}^{2}}-1\right)$

Q 22 . $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}-8$
SOLUTION :
$=(a+b)^{3}-8$ $\left[\because \mathrm{a}^{3}+3 \mathrm{a}^{2} \mathrm{~b}+3 \mathrm{ab}^{2}+\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})^{3}\right]$
$=(a+b)^{3}-2^{3}$
$=(a+b-2)\left((a+b)^{2}+(a+b) \times 2+2^{2}\right)$
$=(a+b-2)\left(a^{2}+2 a b+b^{2}+2 a+2 b+4\right)$
$\therefore a^{3}+3 a^{2} b+3 a b^{2}+b^{3}-8=(a+b-2)\left(a^{2}+2 a b+b^{2}+2 a+2 b+4\right)$

Q 23 . $8 a^{3}-b^{3}-4 a x+2 b x$
SOLUTION :
$=(2 a)^{3}-b^{3}-2 x(2 a-b)$
$=(2 a-b)\left((2 a)^{2}+2 a \times b+b^{2}\right)-2 x(2 a-b)$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(2 a-b)\left(4 a^{2}+2 a b+b^{2}-2 x\right)$
$\therefore 8 a^{3}-b^{3}-4 a x+2 b x=(2 a-b)\left(4 a^{2}+2 a b+b^{2}-2 x\right)$

Q24.i. $\frac{173 \times 173 \times 173+127 \times 127 \times 127}{173 \times 173-173 \times 127+127 \times 127}$
SOLUTION :
$=\frac{173^{3}+127^{3}}{173^{2}-173 \times 127+127^{2}}$
$=\frac{(173+127)\left(173^{2}-173 \times 127+127^{2}\right)}{173^{2}-173 \times 127+127^{2}}$ $\because\left[a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$
$=(173+127)$
$=300$

Q24. ii $\frac{1.2 \times 1.2 \times 1.2-0.2 \times 0.2 \times 0.2}{1.2 \times 1.2+1.2 \times 0.2+0.2 \times 0.2}$
SOLUTION :
$=\frac{1.2^{3}-0.2^{3}}{1.2^{2}+1.2 \times 0.2+0.2^{2}}$
$=\frac{(1.2-0.2)\left((1.2)^{2}+1.2 \times 0.2+(0.2)^{2}\right)}{1.2^{2}+1.2 \times 0.2+0.2^{2}}$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(1.2-0.2)$
$=1.0$

Q24. iii . $\frac{155 \times 155 \times 155-55 \times 55 \times 55}{155 \times 155+155 \times 55+55 \times 55}$
SOLUTION :
$=\frac{155^{3}-55^{3}}{155^{2}+155 \times 55+55^{2}}$
$=\frac{(155-55)\left(155^{2}+155 \times 55+55^{2}\right)}{155^{2}+155 \times 55+55^{2}}$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$
$=(155-55)$
$=100$

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