Exercise 3F 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 3F RS Aggarwal Class 9 contains a total of thirty eight questions. The questions are based on the following topic.

Factorisation of Sum or Difference of Cubes:
(i) $(x^{3} + y^{3}) = (x + y) (x^{2} - x y + y^{2})$
(ii) $(x^{3} - y^{3}) = (x - y) (x^{2} + x y + y^{2})$

Exercise 3F 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 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18 Q.19 Q.20 Q.21 Q.22 Q.23 Q.24 Q.25 Q.26 Q.27 Q.28 Q.29 Q.30 Q.31 Q.32 Q.33 Q.34 Q.35 Q.36 Q.37 Q.38

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

Factorise:

1. $x^{3} + 27$

Answer

We have
$x^{3} + 27$

= $x^{3} + 3^{3}$
Using the identity: $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$
where $a = x, b = 3$

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

= $(x + 3) (x^{2} - 3 x + 9)$ .

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

2. $27 a^{3} + 64 b^{3}$

Answer

We have
$27 a^{3} + 64 b^{3}$

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

= $(3 a + 4 b) [{(3 a)}^{2} - 3 a \cdot 4 b + {(4 b)}^{2}]$

= $(3 a + 4 b) (9 a^{2} - 12 a b + 16 b^{2})$ .

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

3. $125 a^{3} + \frac{1}{8}$

Answer

We have
$125 a^{3} + \frac{1}{8}$

= ${(5 a)}^{3} + {(\frac{1}{2})}^{3}$
Using the identity : $(x^{3} + y^{3}) = (x + y) (x^{2} - x y + y^{2})$
where $x = 5 a, y = \frac{1}{2}$ .

= $(5 a + \frac{1}{2}) [{(5 a)}^{2} - 5 a \cdot \frac{1}{2} + {(\frac{1}{2})}^{2}]$

= $(5 a + \frac{1}{2}) (25 a^{2} - \frac{5 a}{2} + \frac{1}{4})$ .

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

4. $216 x^{3} + \frac{1}{125}$

Answer

We have
$216 x^{3} + \frac{1}{125}$

= ${(6 x)}^{3} + {(\frac{1}{5})}^{3}$
Using the identity: $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$
where $a = 6 x, b = \frac{1}{5}$

= $(6 x + \frac{1}{5}) [{(6 x)}^{2} - 6 x \cdot \frac{1}{5} + {(\frac{1}{5})}^{2}]$

= $(6 x + \frac{1}{5}) (36 x^{2} - \frac{6 x}{5} + \frac{1}{25})$ .

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

5. $16 x^{4} + 54 x$

Answer

We have
$16 x^{4} + 54 x$

= $2 x (8 x^{3} + 27)$

= $2 x [{(2 x)}^{3} + 3^{3}]$
Using the identity : $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$ ,
where $a = 2 x, b = 3$ , we get

= $2 x (2 x + 3) [{(2 x)}^{2} - 2 x \cdot 3 + 3^{2}]$

= $2 x (2 x + 3) (4 x^{2} - 6 x + 9)$ .

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

6. $7 a^{3} + 56 b^{3}$

Answer

We have
$7 a^{3} + 56 b^{3}$

= $7 (a^{3} + 8 b^{3})$

= $7 [a^{3} + {(2 b)}^{3}]$
Using the identity : $(x^{3} + y^{3}) = (x + y) (x^{2} - x y + y^{2})$
where $x = a, y = 2 b$ .

= $7 (a + 2 b) [a^{2} - a \cdot 2 b + {(2 b)}^{2}]$

= $7 (a + 2 b) (a^{2} - 2 a b + 4 b^{2})$ .

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

7. $x^{5} + x^{2}$

Answer

We have
$x^{5} + x^{2}$

= $x^{2} (x^{3} + 1)$

= $x^{2} (x^{3} + 1^{3})$
Using the identity : $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$ , where $a = x, b = 1$ , we get

= $x^{2} (x + 1) [x^{2} - x \cdot 1 + 1^{2}]$

= $x^{2} (x + 1) (x^{2} - x + 1)$ .

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

8. $a^{3} + 0.008$

Answer

We have
$a^{3} + 0.008$

= $a^{3} + {(0.2)}^{3}$ ... $[∵ 0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = {(\frac{2}{10})}^{3} = {(0.2)}^{3}]$

= $(a + 0.2) [a^{2} - a \times 0.2 + {(0.2)}^{2}]$ ... $[∵ (x^{3} + y^{3}) = (x + y) (x^{2} - x y + y^{2})]$

= $(a + 0.2) [a^{2} - 0.2 a + 0.04]$

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

9. $1 - 27 a^{3}$

Answer

We have
$1 - 27 a^{3}$

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

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

= $(1 - 3 a) (1 + 3 a + 9 a^{2})$ .

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

10. $64 a^{3} - 343$

Answer

We have
$64 a^{3} - 343$

= ${(4 a)}^{3} - 7^{3}$

= $(4 a - 7) [{(4 a)}^{2} + 4 a \times 7 + 7^{2}]$ ... $[∵ (x^{3} - y^{3}) = (x - y) (x^{2} + x y + y^{2})]$

= $(4 a - 7) (16 a^{2} + 28 a + 49)$

The question number 11 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

11. $x^{3} - 512$

Answer

We have
$x^{3} - 512$

= $x^{3} - 8^{3}$

= $(x - 8) [x^{2} + x \times 8 + 8^{2}]$ ... $[∵ (a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})]$

= $(x - 8) (x^{2} + 8 x + 64)$ .

The question number 12 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

12. $a^{3} - 0.064$

Answer

We have
$a^{3} - 0.064$

= $a^{3} - {(0.4)}^{3}$

= $(a - 0.4) [a^{2} + a \times 0.4 + {(0.4)}^{2}]$ ... $[∵ (x^{3} - y^{3}) = (x - y) (x^{2} + x y + y^{2})]$

= $(a - 0.4) [a^{2} + 0.4 a + 0.16]$ .

The question number 13 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

13. $8 x^{3} - \frac{1}{27 y^{3}}$

Answer

We have
$8 x^{3} - \frac{1}{27 y^{3}}$

= ${(2 x)}^{3} - {(\frac{1}{3 y})}^{3}$

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

= $(2 x - \frac{1}{3 y}) (4 x^{2} + \frac{2 x}{3 y} + \frac{1}{9 y^{2}})$ .

The question number 14 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

14. $\frac{x^{3}}{216} - 8 y^{3}$

Answer

We have
$\frac{x^{3}}{216} - 8 y^{3}$

= ${(\frac{x}{6})}^{3} - {(2 y)}^{3}$ ...

= $(\frac{x}{6} - 2 y) [{(\frac{x}{6})}^{2} + \frac{x}{6} \times 2 y + {(2 y)}^{2}]$ ... $[∵ (a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})]$

= $(\frac{x}{6} - 2 y) (\frac{x^{2}}{36} + \frac{x y}{3} + 4 y^{2})$ .

The question number 15 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

15. $x - 8 x y^{3}$

Answer

We have
$x - 8 x y^{3}$

= $x (1 - 8 y^{3})$

= $x [1^{3} - {(2 y)}^{3}]$

= $x [(1 - 2 y) {1^{2} + 1 \times 2 y + {(2 y)}^{2}}]$

= $x (1 - 2 y) (1 + 2 y + 4 y^{2})$ .

The question number 16 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

16. $32 x^{4} - 500 x$

Answer

We have
$32 x^{4} - 500 x$

= $4 x {{(2 x)}^{3} - 5^{3}}$

= $4 x [(2 x - 5) {{(2 x)}^{2} + 2 x \times 5 + 5^{2}}]$

= $4 x (2 x - 5) (4 x^{2} + 10 x + 25)$ .

The question number 17 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

17. $3 a^{7} b - 81 a^{4} b^{4}$

Answer

We have
$3 a^{7} b - 81 a^{4} b^{4}$

= $3 a^{4} b (a^{3} - 27 b^{3})$

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

= $3 a^{4} b [(a - 3 b) {a^{2} + a \times 3 b + {(3 b)}^{2}}]$

= $3 a^{4} b [(a - 3 b) {a^{2} + 3 a b + 9 b^{2}}]$

= $3 a^{4} b (a - 3 b) (a^{2} + 3 a b + 9 b^{2})$ .

The question number 18 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

18. $x^{4} y^{4} - x y$

Answer

We have
$x^{4} y^{4} - x y$

= $x y (x^{3} y^{3} - 1)$

= $x y {{(x y)}^{3} - 1^{3}}$

= $x y [(x y - 1) {{(x y)}^{2} + x y \times 1 + 1^{2}}]$

= $x y (x y - 1) (x^{2} y^{2} + x y + 1)$

The question number 19 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

19. $8 x^{2} y^{3} - x^{5}$

Answer

We have
$8 x^{2} y^{3} - x^{5}$

= $x^{2} (8 y^{3} - x^{3})$

= $x^{2} {{(2 y)}^{3} - x^{3}}$

= $x^{2} [(2 y - x) {{(2 y)}^{2} + 2 y \times x + x^{2}}]$

= $x^{2} (2 y - x) (4 y^{2} + 2 x y + x^{2})$ .

The question number 20 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

20. $1029 - 3 x^{3}$

Answer

We have
$1029 - 3 x^{3}$

= $3 (343 - x^{3})$

= $3 (7^{3} - x^{3})$

= $3 (7 - x) (7^{2} + 7 \times x + x^{2})$

= $3 (7 - x) (49 + 7 x + x^{2})$

The question number 21 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

21. $x^{6} - 729$

Answer

We have
$x^{6} - 729$

= ${(x^{3})}^{2} - {(27)}^{2}$
Applying the identity: $(a^{2} - b^{2}) = (a - b) (a + b)$
where, $a = x^{3}, b = 27$ , we get

$(x^{3} - 27) (x^{3} + 27)$

= $(x^{3} - 3^{3}) (x^{3} + 3^{3})$
Now, applying the identities:
$(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$
$(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$
where $a = x, b = 3$ , we get

$[(x - 3) (x^{2} + x \times 3 + 3^{2})] [(x + 3) (x^{2} - x \times 3 + 3^{2})]$

= $(x - 3) (x + 3) (x^{2} + 3 x + 9) (x^{2} - 3 x + 9)$ .

The question number 22 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

22. $x^{9} - y^{9}$

Answer

We have
$x^{9} - y^{9}$

= ${(x^{3})}^{3} - {(y^{3})}^{3}$
Applying the identity: $(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$

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

= $(x^{3} - y^{3}) (x^{6} + x^{3} y^{3} + y^{6})$
Again, applying the identity: $(a^{3} - b^{3}) = (a - b) (a^{2} + a b + b^{2})$

= $(x - y) (x^{2} + x y + y^{2}) (x^{6} + x^{3} y^{3} + y^{6})$ .

The question number 23 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

23. ${(a + b)}^{3} - {(a - b)}^{3}$

Answer

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

Applying the identities:
${(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)$
${(a - b)}^{3} = a^{3} - b^{3} - 3 a b (a + b)$
We get

$[a^{3} + b^{3} + 3 a b (a + b)] - [a^{3} - b^{3} - 3 a b (a - b)]$

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

$= a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} - a^{3} + b^{3} + 3 a^{2} b - 3 a b^{2}$
Eliminating like terms with opposite signs, we get.

= $6 a^{2} b + 2 b^{3}$

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

The question number 24 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

24. $8 a^{3} - b^{3} - 4 a x + 2 b x$

Answer

We have
$8 a^{3} - b^{3} - 4 a x + 2 b x$

= ${(2 a)}^{3} - b^{3} - 2 x (2 a - b)$

= $[{(2 a)}^{3} - b^{3}] - 2 x (2 a - b)$
Applying the identity:
$(x^{3} - y^{3}) = (x - y) (x^{2} + x y + y^{2})$
where $x = 2 a, y = b$

= $[(2 a - b) {{(2 a)}^{2} + 2 a \times b + b^{2}}] - 2 x (2 a - b)$

= $(2 a - b) (4 a^{2} + 2 a b + b^{2}) - 2 x (2 a - b)$

= $(2 a - b) [(4 a^{2} + 2 a b + b^{2}) - 2 x]$

= $(2 a - b) (4 a^{2} + 2 a b + b^{2} - 2 x)$ .

The question number 25 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

25. $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8$

Answer

We have
$a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8$

= $(a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}) - 8$

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

= $[(a + b) - 2] [{(a + b)}^{2} + (a + b) \times 2 + 2^{2}]$

= $(a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 4]$

The question number 26 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

26. $a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

Answer

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

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

= $[(a - \frac{1}{a}) {a^{2} + a \times \frac{1}{a} + {(\frac{1}{a})}^{2}}] - 2 (a - \frac{1}{a})$

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

= $(a - \frac{1}{a}) [(a^{2} + 1 + \frac{1}{a^{2}}) - 2]$

= $(a - \frac{1}{a}) (a^{2} - 1 + \frac{1}{a^{2}})$ .

The question number 27 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

27. $2 a^{3} + 16 b^{3} - 5 a - 10 b$

Answer

We have
$2 a^{3} + 16 b^{3} - 5 a - 10 b$

= $(2 a^{3} + 16 b^{3}) - (5 a + 10 b)$

= $2 (a^{3} + 8 b^{3}) - 5 (a + 2 b)$

= $2 [a^{3} + {(2 b)}^{3}] - 5 (a + 2 b)$

= $2 [(a + 2 b) {a^{2} - a \times 2 b + {(2 b)}^{2}}] - 5 (a + 2 b)$

= $2 [(a + 2 b) (a^{2} - 2 a b + 4 b^{2})] - 5 (a + 2 b)$

= $2 (a + 2 b) (a^{2} - 2 a b + 4 b^{2}) - 5 (a + 2 b)$

= $(a + 2 b) [2 (a^{2} - 2 a b + 4 b^{2}) - 5]$

= $(a + 2 b) [2 a^{2} - 4 a b + 8 b^{2} - 5]$

= $(a + 2 b) (2 a^{2} - 4 a b + 8 b^{2} - 5)$ .

The question number 28 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

28. $a^{6} + b^{6}$

Answer

We have
$a^{6} + b^{6}$

= ${(a^{2})}^{3} + {(b^{2})}^{3}$
Applying the identity: $(x^{3} + y^{3}) = (x + y) (x^{2} - x y + y^{2})$
where $x = a^{2}, y = b^{2}$

= $(a^{2} + b^{2}) [{(a^{2})}^{2} - a^{2} \times b^{2} + {(b^{2})}^{2}]$

= $(a^{2} + b^{2}) [a^{4} - a^{2} b^{2} + b^{4}]$ .

The question number 29 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

29. $a^{12} - b^{12}$

Answer

We have
$a^{12} - b^{12}$

= ${(a^{6})}^{2} - {(b^{6})}^{2}$

= $(a^{6} - b^{6}) (a^{6} + b^{6})$

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

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

$= (a^{3} - b^{3}) (a^{3} + b^{3}) (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$

$= (a - b) (a^{2} + a b + b^{2}) (a + b) (a^{2} - a b + b^{2}) (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$

$= (a - b) (a + b) (a^{2} + b^{2}) (a^{2} + a b + b^{2}) (a^{2} - a b + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$

The question number 30 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

30. $x^{6} - 7 x^{3} - 8$

Answer

First Method:

We have
$x^{6} - 7 x^{3} - 8$

$= x^{6} - 7 x^{3} - 7 - 1$

$= x^{6} - 1 - 7 x^{3} - 7$

$= (x^{6} - 1) - 7 (x^{3} + 1)$

$= [{(x^{3})}^{2} - 1^{3}] - 7 (x^{3} + 1)$

$= [(x^{3} - 1) (x^{3} + 1)] - 7 (x^{3} + 1)$

$= (x^{3} - 1) (x^{3} + 1) - 7 (x^{3} + 1)$

$= (x^{3} + 1) [(x^{3} - 1) - 7]$

$= (x^{3} + 1) [x^{3} - 1 - 7]$

$= (x^{3} + 1) [x^{3} - 8]$

$= (x^{3} + 1) [x^{3} - 2^{3}]$

$= (x + 1) (x^{2} - x \times 1 + 1^{2}) [(x - 2) (x^{2} + 2 \times x + 2^{2})]$

$= (x + 1) (x^{2} - x + 1) (x - 2) (x^{2} + 2 x + 4)$

$= (x + 1) (x - 2) (x^{2} - x + 1) (x^{2} + 2 x + 4)$ .

Second Method:

We have
$x^{6} - 7 x^{3} - 8$

Let $x^{3} = y$ , the given polynomial becomes
$y^{2} - 7 y - 8$

$= y^{2} - 8 y + y - 8$

$= y (y - 8) + 1 (y - 8)$

$= (y + 1) (y - 8)$

$= (x^{3} + 1) (x^{3} - 8)$ ... [putting $y = x^{3}$ ]

$= (x^{3} + 1^{3}) (x^{3} - 2^{3})$

$= (x + 1) (x^{2} - x \times 1 + 1^{2}) (x - 2) (x^{2} + x \times 2 + 2^{2})$

$= (x + 1) (x^{2} - x + 1) (x - 2) (x^{2} + 2 x + 4)$

= $(x + 1) (x - 2) (x^{2} - x + 1) (x^{2} + 2 x + 4)$ .

The question number 31 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

31. $x^{3} - 3 x^{2} + 3 x + 7$

Answer

First Method:

We have
$x^{3} - 3 x^{2} + 3 x + 7$

Adding and subtracting 1, we get
$x^{3} - 3 x^{2} + 3 x - 1 + 1 + 7$

= $(x^{3} - 3 x^{2} + 3 x - 1) + (1 + 7)$

= $(x^{3} - 3 \cdot x^{2} \cdot 1 + 3 \cdot x \cdot 1^{2} - 1) + 8$
Using the identity : ${(a - b)}^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ , where $a = x, b = 1$ , we get

= ${(x - 1)}^{3} + 2^{3}$
Again, using the identity: $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
where $a = x - 1, b = 2$ , we get

= ${(x - 1) + 2} [{(x - 1)}^{2} - (x - 1) \times 2 + 2^{2}]$

= $(x + 1) [{(x - 1)}^{2} - 2 (x - 1) + 4]$

= $(x + 1) [x^{2} - 2 x + 1 - 2 x + 2 + 4]$

= $(x + 1) (x^{2} - 4 x + 7)$ .

Second Method:

Let $p (x) = x^{3} - 3 x^{2} + 3 x + 7$ ,
∵ $p (x)$ is a cubic polynomial, we will find one of its factors by finding its one zero.

To find a zero, we find all factors of the constant term of $p (x)$ .

All factors of 7 are ±1, ±7.

Now, we will check if any of the values $p (1), p (- 1), p (7), p (- 7)$ is zero.

$∵ p (- 1) = {(- 1)}^{3} - 3 {(- 1)}^{2} + 3 (- 1) + 7$
$= - 1 - 3 - 3 + 7$
$= - 7 + 7$
$= 0$

$∴ - 1$ is a zero of $p (x)$ .

According to factor theorem, $(x + 1)$ is a factor of $p (x)$ .
It implies that $p (x)$ when divided by $(x + 1)$ gives zero remainder.

After division of $p (x)$ by $(x + 1)$ , we get $(x^{2} - 4 x + 7)$ as quotient and 0 as remainder.

Hence, $x^{3} - 3 x^{2} + 3 x + 7$ = $(x + 1) (x^{2} - 4 x + 7)$ .

The question number 32 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

32. ${(x + 1)}^{3} + {(x - 1)}^{3}$

Answer

We have
${(x + 1)}^{3} + {(x - 1)}^{3}$
Using the identity : $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
where $a = (x + 1), b = (x - 1)$ , we get

$= {(x + 1) + (x - 1)} [{(x + 1)}^{2} - (x + 1) (x - 1) + {(x - 1)}^{2}]$

$= (x + 1 + x - 1) [x^{2} + 2 x + 1^{2} - (x^{2} - 1) + x^{2} - 2 x + 1^{2}]$

$= 2 x [x^{2} + 2 x + 1 - x^{2} + 1 + x^{2} - 2 x + 1^{2}]$
Cancelling like terms with opposite signs, we get

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

The question number 33 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

33. ${(2 a + 1)}^{3} + {(a - 1)}^{3}$

Answer

We have
${(2 a + 1)}^{3} + {(a - 1)}^{3}$
Using the identity : $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$
where $x = (2 a + 1), y = (a - 1)$ , we get

$= {(2 a + 1) + (a - 1)} [{(2 a + 1)}^{2} - (2 a + 1) (a - 1) + {(a - 1)}^{2}]$

$= (2 a + 1 + a - 1) [{(2 a)}^{2} + 2 \cdot (2 a) . 1 + 1^{2} - {2 a (a - 1) + 1 (a - 1)} + a^{2} - 2 \cdot a \cdot 1 + 1^{2}]$

$= 3 a [4 a^{2} + 4 a + 1 - {2 a^{2} - 2 a + a - 1} + a^{2} - 2 a + 1]$

$= 3 a [4 a^{2} + 4 a + 1 - 2 a^{2} + 2 a - a + 1 + a^{2} - 2 a + 1]$

$= 3 a [4 a^{2} - 2 a^{2} + a^{2} + 4 a + 2 a - a - 2 a + 1 + 1 + 1]$

$= 3 a [3 a^{2} + 3 a + 3]$

$= 3 a \times 3 [a^{2} + a + 1]$

$= 9 a (a^{2} + a + 1)$ .

The question number 34 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

34. $8 {(x + y)}^{3} - 27 {(x - y)}^{3}$

Answer

We have
$8 {(x + y)}^{3} - 27 {(x - y)}^{3}$

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

$= [2 (x + y) - 3 (x - y)] [{2 (x + y)}^{2} + {2 (x + y)} \times {3 (x - y)} + {3 (x - y)}^{2}]$

$= [2 x + 2 y - 3 x + 3 y] [2^{2} {(x + y)}^{2} + 2 \times 3 \times (x + y) \times (x - y) + 3^{2} {(x - y)}^{2}]$

$= (- x + 5 y) [4 {x^{2} + 2 x y + y^{2}} + 6 (x^{2} - y^{2}) + 9 {x^{2} - 2 x y + y^{2}}]$

$= (- x + 5 y) [4 x^{2} + 8 x y + 4 y^{2} + 6 x^{2} - 6 y^{2} + 9 x^{2} - 18 x y + 9 y^{2}]$

$= (- x + 5 y) [4 x^{2} + 6 x^{2} + 9 x^{2} - 18 x y + 8 x y + 4 y^{2} - 6 y^{2} + 9 y^{2}]$

$= (- x + 5 y) (19 x^{2} - 10 x y + 7 y^{2})$

The question number 35 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

35. ${(x + 2)}^{3} + {(x - 2)}^{3}$

Answer

We have
${(x + 2)}^{3} + {(x - 2)}^{3}$
Applying the identity : $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
where $a = (x + 2), b = (x - 2)$ , we get

$= {(x + 2) + (x - 2)} [{(x + 2)}^{2} - (x + 2) (x - 2) + {(x - 2)}^{2}]$

$= (x + 2 + x - 2) [x^{2} + 2 \cdot 2 \cdot x + 2^{2} - (x^{2} - 2^{2}) + x^{2} - 2 \cdot 2 \cdot x + 2^{2}]$

$= 2 x [x^{2} + 4 x + 4 - x^{2} + 4 + x^{2} - 4 x + 4]$

$= 2 x [x^{2} + 4 + 4 + 4]$

$= 2 x (x^{2} + 12)$ .

The question number 36 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

36. ${(x + 2)}^{3} - {(x - 2)}^{3}$

Answer

We have
${(x + 2)}^{3} - {(x - 2)}^{3}$
Applying the identity : $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
where $a = (x + 2), b = (x - 2)$ , we get

$= {(x + 2) - (x - 2)} [{(x + 2)}^{2} + (x + 2) (x - 2) + {(x - 2)}^{2}]$

$= (x + 2 - x + 2) [x^{2} + 2 \cdot 2 \cdot x + 2^{2} + (x^{2} - 2^{2}) + x^{2} - 2 \cdot 2 \cdot x + 2^{2}]$

$= 4 [x^{2} + 4 x + 4 + x^{2} - 4 + x^{2} - 4 x + 4]$

$= 4 [x^{2} + 4 + x^{2} + x^{2}]$

$= 4 (3 x^{2} + 4)$ .

The question number 37 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

37. Prove that $\frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = 1$ .

Answer

To Prove: $\frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = 1$ .

Taking LHS, we get
$\frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15}$

= $\frac{{(0.85)}^{3} + {(0.15)}^{3}}{{(0.85)}^{2} - 0.85 \times 0. 15 + {(0.15)}^{2}}$

= $\frac{(0.85 + 0.15) {{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}}}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}}$
Cancelling the same factors in numerator and denominator, we get

= $(0.85 + 0.15)$

= 1 = RHS
Hence proved.

The question number 38 of Exercise 3F RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

38. Prove that $\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = 50$ .

Answer

To Prove: $\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = 50$ .

Taking LHS, we get
$\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9}$

= $\frac{{(59)}^{3} - 9^{3}}{{(59)}^{2} + 59 \times 9 + 9^{2}}$

= $\frac{(59 - 9) {{(59)}^{2} - 59 \times 9 + 9^{2}}}{{(59)}^{2} - 59 \times 9 + 9^{2}}$
Cancelling the same factors in numerator and denominator, we get

= $(59 - 9)$

= $50$ = RHS
Hence, proved.

Exercise 3F RS Aggarwal Class 9 Mathematics Solutions

Factorise:

1. x3+27

Answer

2. 27a3+64b3

Answer

3. 125a3+18

Answer

4. 216x3+1125

Answer

5. 16x4+54x

Answer

6. 7a3+56b3

Answer

7. x5+x2

Answer

8. a3+0.008

Answer

9. 1−27a3

Answer

10. 64a3−343

Answer

11. x3−512

Answer

12. a3−0.064

Answer

13. 8x3−127y3

Answer

14. x3216−8y3

Answer

15. x−8xy3

Answer

16. 32x4−500x

Answer

17. 3a7b−81a4b4

Answer

18. x4y4−xy

Answer

19. 8x2y3−x5

Answer

20. 1029−3x3

Answer

21. x6−729

Answer

22. x9−y9

Answer

23. (a+b)3−(a−b)3

Answer

24. 8a3−b3−4ax+2bx

Answer

25. a3+3a2b+3ab2+b3−8

Answer

26. a3−1a3−2a+2a

Answer

27. 2a3+16b3−5a−10b

Answer

28. a6+b6

Answer

29. a12−b12

Answer

30. x6−7x3−8

Answer

31. x3−3x2+3x+7

Answer

32. (x+1)3+(x−1)3

Answer

33. (2a+1)3+(a−1)3

Answer

34. 8(x+y)3−27(x−y)3

Answer

35. (x+2)3+(x−2)3

Answer

36. (x+2)3−(x−2)3

Answer

37. Prove that 0.85×0.85×0.85+0.15×0.15×0.150.85×0.85−0.85×0.15+0.15×0.15=1.

Answer

38. Prove that 59×59×59−9×9×959×59+59×9+9×9=50.

Answer

Related Links 🔗

Important Links

1. $x^{3} + 27$

2. $27 a^{3} + 64 b^{3}$

3. $125 a^{3} + \frac{1}{8}$

4. $216 x^{3} + \frac{1}{125}$

5. $16 x^{4} + 54 x$

6. $7 a^{3} + 56 b^{3}$

7. $x^{5} + x^{2}$

8. $a^{3} + 0.008$

9. $1 - 27 a^{3}$

10. $64 a^{3} - 343$

11. $x^{3} - 512$

12. $a^{3} - 0.064$

13. $8 x^{3} - \frac{1}{27 y^{3}}$

14. $\frac{x^{3}}{216} - 8 y^{3}$

15. $x - 8 x y^{3}$

16. $32 x^{4} - 500 x$

17. $3 a^{7} b - 81 a^{4} b^{4}$

18. $x^{4} y^{4} - x y$

19. $8 x^{2} y^{3} - x^{5}$

20. $1029 - 3 x^{3}$

21. $x^{6} - 729$

22. $x^{9} - y^{9}$

23. ${(a + b)}^{3} - {(a - b)}^{3}$

24. $8 a^{3} - b^{3} - 4 a x + 2 b x$

25. $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8$

26. $a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

27. $2 a^{3} + 16 b^{3} - 5 a - 10 b$

28. $a^{6} + b^{6}$

29. $a^{12} - b^{12}$

30. $x^{6} - 7 x^{3} - 8$

31. $x^{3} - 3 x^{2} + 3 x + 7$

32. ${(x + 1)}^{3} + {(x - 1)}^{3}$

33. ${(2 a + 1)}^{3} + {(a - 1)}^{3}$

34. $8 {(x + y)}^{3} - 27 {(x - y)}^{3}$

35. ${(x + 2)}^{3} + {(x - 2)}^{3}$

36. ${(x + 2)}^{3} - {(x - 2)}^{3}$

37. Prove that $\frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = 1$ .

38. Prove that $\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = 50$ .