Exercise 3E RS Aggarwal Class 9 Factorisation of Polynomials Best Solutions

Exercise 1A Exercise 1B Exercise 1C Exercise 1D Exercise 1E Exercise 1F Exercise 1G Exercise 2A Exercise 2B Exercise 2C Exercise 2D Exercise 3A Exercise 3B Exercise 3C Exercise 3D Exercise 3E Exercise 3F

Exercise 3E RS Aggarwal Class 9 contains solutions of all the ten questions. The questions are based on the following topic.

Cube of a Binomial:
(i) ${(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)$
(ii) ${(x - y)}^{3} = x^{3} + y^{3} - 3 x y (x - y)$

Exercise 3E RS Aggarwal Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10

The question number 1 of Exercise 3E RS Aggarwal Class 9 asks us to expand the given algebraic expression.

**1. Expand:**

(i). ${(3 x + 2)}^{3}$

Answer

We have
${(3 x + 2)}^{3}$

= ${(3 x)}^{3} + 2^{3} + 3 \cdot 3 x \cdot 2 \cdot (3 x + 2)$ ... $[∵ {(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)]$

= $3^{3} \cdot x^{3} + 2^{3} + 18 x \cdot (3 x + 2)$

= $27 x^{3} + 8 + 54 x^{2} + 36 x$ .

(ii). ${(3 a + \frac{1}{4 b})}^{3}$

Answer

We have
${(3 a + \frac{1}{4 b})}^{3}$

= ${(3 a)}^{3} + {(\frac{1}{4 b})}^{3} + 3 \cdot 3 a \cdot \frac{1}{4 b} \cdot (3 a + \frac{1}{4 b})$ ... $[∵ {(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)]$

= $3^{3} a^{3} + \frac{1^{3}}{{(4 b)}^{3}} + \frac{9 a}{4 b} \cdot (3 a + \frac{1}{4 b})$

= $27 a^{3} + \frac{1}{64 b^{3}} + \frac{9 a}{4 b} \cdot 3 a + \frac{9 a}{4 b} \cdot \frac{1}{4 b}$

= $27 a^{3} + \frac{1}{64 b^{3}} + \frac{27 a^{2}}{4 b} + \frac{9 a}{16 b^{2}}$

(ii). ${(1 + \frac{2}{3} a)}^{3}$

Answer

We have
${(1 + \frac{2}{3} a)}^{3}$

= $1^{3} + {(\frac{2}{3} a)}^{3} + 3 \cdot 1 \cdot \frac{2}{3} a \cdot (1 + \frac{2}{3} a)$ ... $[∵ {(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)]$

= $1 + \frac{2^{3}}{3^{3}} a^{3} + 2 a \cdot (1 + \frac{2}{3} a)$

= $1 + \frac{8}{27} a^{3} + 2 a \cdot 1 + 2 a \cdot \frac{2}{3} a$

= $1 + \frac{8}{27} a^{3} + 2 a + \frac{4 a^{2}}{3}$ .

The question number 2 of Exercise 3E RS Aggarwal Class 9 asks us to expand the given algebraic expression.

2. Expand

(i). ${(5 a - 3 b)}^{3}$

Answer

We have
${(5 a - 3 b)}^{3}$

= ${(5 a)}^{3} - {(3 b)}^{3} - 3 \cdot 5 a \cdot 3 b \cdot (5 a - 3 b)$ ... $[∵ {(x - y)}^{3} = x^{3} + y^{3} - 3 x y (x - y)]$

= $5^{3} \cdot a^{3} - 3^{3} \cdot b^{3} - 45 a b \cdot (5 a - 3 b)$

= $125 a^{3} - 27 b^{3} - 45 a b \cdot 5 a + 45 a b \cdot 3 b$

= $125 a^{3} - 27 b^{3} - 225 a^{2} b + 135 a b^{2}$ .

(ii). ${(3 x - \frac{5}{x})}^{3}$

Answer

We have
${(3 x - \frac{5}{x})}^{3}$

= ${(3 x)}^{3} - {(\frac{5}{x})}^{3} - 3 \cdot 3 x \cdot \frac{5}{x} \cdot (3 x - \frac{5}{x})$ ... $[∵ {(x - y)}^{3} = x^{3} + y^{3} - 3 x y (x - y)]$

= $3^{3} \cdot x^{3} - \frac{5^{3}}{x^{3}} - 45 \cdot (3 x - \frac{5}{x})$

= $3^{3} \cdot x^{3} - \frac{5^{3}}{x^{3}} - 135 x + 225$

= $27 x^{3} - \frac{125}{x^{3}} - 135 x + \frac{225}{x}$ .

(iii). ${(\frac{4}{5} a - 2)}^{3}$

Answer

We have
${(\frac{4}{5} a - 2)}^{3}$

= ${(\frac{4}{5} a)}^{3} - 2^{3} - 3 \cdot \frac{4}{5} a \cdot 2 \cdot (\frac{4}{5} a - 2)$ ... $[∵ {(x - y)}^{3} = x^{3} + y^{3} - 3 x y (x - y)]$

= $\frac{4^{3}}{5^{3}} a^{3} - 2^{3} - 3 \cdot \frac{4}{5} a \cdot 2 \cdot (\frac{4}{5} a - 2)$

= $\frac{64}{125} a^{3} - 8 - \frac{24 a}{5} \cdot (\frac{4}{5} a - 2)$

= $\frac{64}{125} a^{3} - 8 - \frac{24 a}{5} \cdot \frac{4}{5} a + \frac{24 a}{5} \cdot 2$

= $\frac{64}{125} a^{3} - 8 - \frac{96}{25} a^{2} + \frac{48}{5} a$ .

The question number 3 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

Factorise

3. $8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2}$

Answer

We have
$8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2}$

= ${(2 a)}^{3} + {(3 b)}^{3} + 3 \times {(2 a)}^{2} \times 3 b + 3 \times (2 a) \times {(3 b)}^{2}$
Using $x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2} = {(x + y)}^{3}$ , where $x = 2 a, y = 3 b$ , we get

= ${(2 a + 3 b)}^{3}$

= $(2 a + 3 b) (2 a + 3 b) (2 a + 3 b)$ .

The question number 4 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

4. $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

Answer

We have
$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

= ${(4 a)}^{3} - {(3 b)}^{3} - 3 \times {(4 a)}^{2} \times 3 b + 3 \times (4 a) \times {(3 b)}^{2}$
Using $x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2} = {(x - y)}^{3}$ , where $x = 4 a, y = 3 b$ , we get

= ${(4 a - 3 b)}^{3}$

= $(4 a - 3 b) (4 a - 3 b) (4 a - 3 b)$ .

The question number 5 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

5. $1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{25}$

Answer

We have
$1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{25}$

= $1^{3} + \frac{3^{3}}{5^{3}} a^{3} + 3 \times 1^{2} \times \frac{3}{5} a + 3 \times 1 \times \frac{3^{2}}{5^{2}} a$

= $1^{3} + {(\frac{3}{5} a)}^{3} + 3 \times 1^{2} \times \frac{3}{5} a + 3 \times 1 \times {(\frac{3}{5} a)}^{2}$
Using identity $x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2} = {(x + y)}^{3}$ , where $x = 1, y = \frac{3}{5} a$ , we get

= ${(1 + \frac{3}{5} a)}^{3}$

= $(1 + \frac{3}{5} a) (1 + \frac{3}{5} a) (1 + \frac{3}{5} a)$ .

The question number 6 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

6. $125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2}$

Answer

We have
$125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2}$

= $5^{3} \cdot x^{3} - 3^{3} \cdot y^{3} - 3 \cdot 5^{2} \cdot x^{2} \cdot 3 y + 3 \cdot 5 x \times 3^{2} \cdot y^{2}$

= ${(5 x)}^{3} - {(3 y)}^{3} - 3 \cdot {(5 x)}^{2} \cdot (3 y) + 3 \cdot (5 x) \cdot {(3 y)}^{2}$

= ${(5 x)}^{3} - {(3 y)}^{3} - 3 \cdot {(5 x)}^{2} \cdot (3 y) + 3 \cdot (5 x) \cdot {(3 y)}^{2}$
Using the identity $x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2} = {(x - y)}^{3}$ , we get

= ${(5 x - 3 y)}^{3}$

= $(5 x - 3 y) (5 x - 3 y) (5 x - 3 y)$ .

The question number 7 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

7. $a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3}$

Answer

We have
$a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3}$

= ${(a x)}^{3} - 3 \cdot {(a x)}^{2} \cdot b + 3 \cdot (a x) \cdot b^{2} - b^{3}$
Using the identity $x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3} = {(x - y)}^{3}$ , we get

= ${(a x - b)}^{3}$

= $(a x - b) (a x - b) (a x - b)$

The question number 8 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

8. $\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8$

Answer

We have
$\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8$

= $\frac{4^{3}}{5^{3}} \cdot a^{3} - 3 \cdot \frac{4^{2}}{5^{2}} a^{2} \cdot 2 + 3 \cdot \frac{4}{5} a \cdot 2^{2} - 2^{3}$

= ${(\frac{4}{5} a)}^{3} - 3 \cdot {(\frac{4}{5} a)}^{2} \cdot 2 + 3 \cdot (\frac{4}{5} a) \cdot 2^{2} - 2^{3}$
Using the identity $x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3} = {(x - y)}^{3}$ , where $x = \frac{4}{5} a, y = 2$ , we get

= ${(\frac{4}{5} a - 2)}^{3}$

= $(\frac{4}{5} a - 2) (\frac{4}{5} a - 2) (\frac{4}{5} a - 2)$

The question number 9 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

9. $a^{3} - 12 a (a - 4) - 64$

Answer

We have
$a^{3} - 12 a (a - 4) - 64$

= $a^{3} - 12 a^{2} + 48 a - 64$

= $a^{3} - 3 \cdot a^{2} \cdot 4 + 3 \cdot a \cdot 4^{2} - 4^{3}$
Using the identity $x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3} = {(x - y)}^{3}$ , where $x = a, y = 4$ .

= ${(a - 4)}^{3}$

= $(a - 4) (a - 4) (a - 4)$ .

The question number 10 of Exercise 3E RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

10. Evaluate

(i) ${(103)}^{3}$

Answer

We have
${(103)}^{3}$

= ${(100 + 3)}^{3}$
Applying the identity : ${(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)$

= ${(100)}^{3} + 3^{3} + 3 \cdot {(100)}^{2} \cdot 3 + 3 \cdot 100 \cdot 3^{2}$

= $1000000 + 27 + 90000 + 2700$

= $1092727$ .

(ii) ${(99)}^{3}$

Answer

We have
${(99)}^{3}$

= ${(100 - 1)}^{3}$

= ${(100 - 1)}^{3}$
Applying the identity : ${(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)$

= ${(100)}^{3} - 1^{3} - 3 \cdot {(100)}^{2} \cdot 1 + 3 \cdot 100 \cdot 1^{2}$

= $1000000 - 1 - 30000 + 300$

= $1000300 - 30001$

= $970299$ .

Exercise 3E RS Aggarwal Class 9 Mathematics Solutions

1. Expand:

(i). (3x+2)3

Answer

(ii). (3a+14b)3

Answer

(ii). (1+23a)3

Answer

2. Expand

(i). (5a−3b)3

Answer

(ii). (3x−5x)3

Answer

(iii). (45a−2)3

Answer

Factorise

3. 8a3+27b3+36a2b+54ab2

Answer

4. 64a3−27b3−144a2b+108ab2

Answer

5. 1+27125a3+9a5+27a225

Answer

6. 125x3−27y3−225x2y+135xy2

Answer

7. a3x3−3a2bx2+3ab2x−b3

Answer

8. 64125a3−9625a2+485a−8

Answer

9. a3−12a(a−4)−64

Answer

10. Evaluate

(i) (103)3

Answer

(ii) (99)3

Answer

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Important Links

**1. Expand:**

(i). ${(3 x + 2)}^{3}$

(ii). ${(3 a + \frac{1}{4 b})}^{3}$

(ii). ${(1 + \frac{2}{3} a)}^{3}$

(i). ${(5 a - 3 b)}^{3}$

(ii). ${(3 x - \frac{5}{x})}^{3}$

(iii). ${(\frac{4}{5} a - 2)}^{3}$

3. $8 a^{3} + 27 b^{3} + 36 a^{2} b + 54 a b^{2}$

4. $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$

5. $1 + \frac{27}{125} a^{3} + \frac{9 a}{5} + \frac{27 a^{2}}{25}$

6. $125 x^{3} - 27 y^{3} - 225 x^{2} y + 135 x y^{2}$

7. $a^{3} x^{3} - 3 a^{2} b x^{2} + 3 a b^{2} x - b^{3}$

8. $\frac{64}{125} a^{3} - \frac{96}{25} a^{2} + \frac{48}{5} a - 8$

9. $a^{3} - 12 a (a - 4) - 64$

(i) ${(103)}^{3}$

(ii) ${(99)}^{3}$