myhelper: RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3F

EXERCISE 3F

PAGE NO-129

Question 1:

Factorize:
x³ + 27

Answer 1:

$x^{3} + 27 = {(x)}^{3} + {(3)}^{3} = (x + 3) (x^{2} - 3 x + 3^{2}) = (x + 3) (x^{2} - 3 x + 9)$

Question 2:

Factorise
27a³ + 64b³

Answer 2:

We know that
$x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$
Given: 27a³ + 64b³
x = 3a, y = 4b
$27 a^{3} + 64 b^{3} = (3 a + 4 b) (9 a^{2} + 16 b^{2} - 12 a b)$

Question 3:

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

Answer 3:

$125 a^{3} + \frac{1}{8} = {(5 a)}^{3} + {(\frac{1}{2})}^{3} = (5 a + \frac{1}{2}) [{(5 a)}^{2} - 5 a \times \frac{1}{2} + {(\frac{1}{2})}^{2}] = (5 a + \frac{1}{2}) (25 a^{2} - \frac{5 a}{2} + \frac{1}{4})$

Question 4:

Factorize:
$216 x^{3} + \frac{1}{125}$

Answer 4:

$216 x^{3} + \frac{1}{125} = {(6 x)}^{3} + {(\frac{1}{5})}^{3} = (6 x + \frac{1}{5}) [{(6 x)}^{2} - 6 x \times \frac{1}{5} + {(\frac{1}{5})}^{2}] = (6 x + \frac{1}{5}) (36 x^{2} - \frac{6 x}{5} + \frac{1}{25})$

Question 5:

Factorize:
16x⁴ + 54x

Answer 5:

$16 x^{4} + 54 x = 2 x (8 x^{3} + 27) = 2 x [{(2 x)}^{3} + {(3)}^{3}] = 2 x (2 x + 3) [{(2 x)}^{2} - 2 x \times 3 + 3^{2}] = 2 x (2 x + 3) (4 x^{2} - 6 x + 9)$

Question 6:

Factorize:
7a³ + 56b³

Answer 6:

$7 a^{3} + 56 b^{3} = 7 (a^{3} + 8 b^{3}) = 7 [{(a)}^{3} + {(2 b)}^{3}] = 7 (a + 2 b) [a^{2} - a \times 2 b + {(2 b)}^{2}] = 7 (a + 2 b) (a^{2} - 2 a b + 4 b^{2})$

Question 7:

Factorize:
x⁵ + x²

Answer 7:

$x^{5} + x^{2} = x^{2} (x^{3} + 1) = x^{2} (x^{3} + 1^{3}) = x^{2} (x + 1) (x^{2} - x \times 1 + 1^{2}) = x^{2} (x + 1) (x^{2} - x + 1)$

Question 8:

Factorize:
a³ + 0.008

Answer 8:

$a^{3} + 0.008 = a^{3} + {(0.2)}^{3} = (a + 0.2) [a^{2} - a \times (0.2) + {(0.2)}^{2}] = (a + 0.2) (a^{2} - 0.2 a + 0.04)$

Question 9:

Factorise
1 – 27a³

Answer 9:

$1 - 27 a^{3} = 1^{3} - {(3 a)}^{3} = (1 - 3 a) [1^{2} + 1 \times 3 x + {(3 a)}^{2}] = (1 - 3 a) (1 + 3 a + 9 a^{2})$

Question 10:

Factorize:
64a³ − 343

Answer 10:

$64 a^{3} - 343 = {(4 a)}^{3} - {(7)}^{3} = (4 a - 7) (16 a^{2} + 4 a \times 7 + 7^{2}) = (4 a - 7) (16 a^{2} + 28 a + 49)$

Question 11:

Factorize:
x³ − 512

Answer 11:

$x^{3} - 512 = x^{3} - 8^{3} = (x - 8) (x^{2} + 8 x + 8^{2}) = (x - 8) (x^{2} + 8 x + 64)$

Question 12:

Factorize:
a³ − 0.064

Answer 12:

$a^{3} - 0.064 = {(a)}^{3} - {(0.4)}^{3} = (a - 0.4) [a^{2} + a \times (0.4) + {(0.4)}^{2}] = (a - 0.4) (a^{2} + 0.4 a + 0.16)$

Question 13:

Factorize:
$8 x^{3} - \frac{1}{27 y^{3}}$

Answer 13:

$8 x^{3} - \frac{1}{27 y^{3}} = {(2 x)}^{3} - {(\frac{1}{3 y})}^{3} = (2 x - \frac{1}{3 y}) [{(2 x)}^{2} + 2 x \times \frac{1}{3 y} + {(\frac{1}{3 y})}^{2}] = (2 x - \frac{1}{3 y}) (4 x^{2} + \frac{2 x}{3 y} + \frac{1}{9 y^{2}})$

Question 14:

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

Answer 14:

We know
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
We have,
$\frac{x^{3}}{216} - 8 y^{3} = {(\frac{x}{6})}^{3} - {(2 y)}^{3}$
So, $a = \frac{x}{6}, b = 2 y$
$\frac{x^{3}}{216} - 8 y^{3} = (\frac{x}{6} - 2 y) ({(\frac{x}{6})}^{2} + \frac{x}{6} \times 2 y + {(2 y)}^{2}) = (\frac{x}{6} - 2 y) (\frac{x^{2}}{36} + \frac{x y}{3} + 4 y^{2})$

Question 15:

Factorize:
x − 8xy³

Answer 15:

$x - 8 x y^{3} = x (1 - 8 y^{3}) = x [1^{3} - {(2 y)}^{3}] = x (1 - 2 y) (1^{2} + 1 \times 2 y + {(2 y)}^{2}) = x (1 - 2 y) (1 + 2 y + 4 y^{2})$

Question 16:

Factorise
32x⁴ – 500x

Answer 16:

$32 x^{4} - 500 x = 4 x (8 x^{3} - 125) = 4 x ({(2 x)}^{3} - 5^{3}) we know a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b) a = 2 x, b = 5 32 x^{4} - 500 x = 4 x ({(2 x)}^{3} - 5^{3}) = 4 x (2 x - 5) (4 x^{2} + 25 + 10 x)$

Question 17:

Factorize:
3a⁷b − 81a⁴b⁴

Answer 17:

$3 a^{7} b - 81 a^{4} b^{4} = 3 a^{4} b (a^{3} - 27 b^{3}) = 3 a^{4} b [a^{3} - {(3 b)}^{3}] = 3 a^{4} b (a - 3 b) [a^{2} + a \times 3 b + {(3 b)}^{2}] = 3 a^{4} b (a - 3 b) (a^{2} + 3 a b + 9 b^{2})$

Question 18:

Factorise
x⁴ y⁴ – xy

Answer 18:

Using the identity
$a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$
$x^{4} y^{4} - x y = x y (x^{3} y^{3} - 1) = x y (x y - 1) (x^{2} y^{2} + 1 + x y)$

Question 19:

Factorise
8x² y³ – x⁵

Answer 19:

$8 x^{2} y^{3} - x^{5} = x^{2} (8 y^{3} - x^{3}) = x^{2} (2 y - x) (4 y^{2} + x^{2} + 2 x y)$

Question 20:

Factorise
1029 – 3x³

Answer 20:

$1029 - 3 x^{3} = 3 (343 - x^{3}) = 3 (7^{3} - x^{3}) = 3 (7 - x) (49 + x^{2} + 7 x)$

Question 21:

Factorize:
x⁶ − 729

Answer 21:

$x^{6} - 729 = {(x^{2})}^{3} - {(9)}^{3} = [x^{2} - 9] [{(x^{2})}^{2} + x^{2} \times 9 + 9^{2}] = [x^{2} - 3^{2}] (x^{4} + 9 x^{2} + 81) = (x + 3) (x - 3) (x^{4} + 18 x^{2} + 81 - 9 x^{2}) = (x + 3) (x - 3) [{(x^{2})}^{2} + 2 \times x^{2} \times 9 + 9^{2} - 9 x^{2}] = (x + 3) (x - 3) [{(x^{2} + 9)}^{2} - {(3 x)}^{2}] = (x + 3) (x - 3) (x^{2} + 9 + 3 x) (x^{2} + 9 - 3 x) = (x + 3) (x - 3) (x^{2} + 3 x + 9) (x^{2} - 3 x + 9)$

Question 22:

Factorise
x⁹ – y⁹

Answer 22:

$x^{9} - y^{9} = {(x^{3})}^{3} - {(y^{3})}^{3} we know a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b) a = x^{3}, b = y^{3} So, x^{9} - y^{9} = {(x^{3})}^{3} - {(y^{3})}^{3} = (x^{3} - y^{3}) (x^{6} + y^{6} + x^{3} y^{3}) = (x - y) (x^{2} + y^{2} + x y) (x^{6} + y^{6} + x^{3} y^{3})$

Question 23:

Factorize:
(a + b)³ − (a − b)³

Answer 23:

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

Question 24:

Factorize:
8a³ − b³ − 4ax + 2bx

Answer 24:

$8 a^{3} - b^{3} - 4 a x + 2 b x = [{(2 a)}^{3} - {(b)}^{3}] - 2 x (2 a - b) = (2 a - b) [{(2 a)}^{2} + 2 a b + b^{2}] - 2 x (2 a - b) = (2 a - b) (4 a^{2} + 2 a b + b^{2}) - 2 x (2 a - b) = (2 a - b) (4 a^{2} + 2 a b + b^{2} - 2 x)$

Question 25:

Factorize:
a³ + 3a²b + 3ab² + b³ − 8

Answer 25:

$a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8 = (a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}) - 8 = [a^{3} + b^{3} + 3 a b (a + b)] - 8 = {(a + b)}^{3} - 2^{3} = (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 2^{2}] = (a + b - 2) [{(a + b)}^{2} + 2 (a + b) + 4]$

Question 26:

Factorize:
$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a}$

Answer 26:

$a^{3} - \frac{1}{a^{3}} - 2 a + \frac{2}{a} = (a^{3} - \frac{1}{a^{3}}) - 2 (a - \frac{1}{a}) = [{(a)}^{3} - {(\frac{1}{a})}^{3}] - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) [a^{2} + a \times \frac{1}{a} + {(\frac{1}{a})}^{2}] - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a^{2}}) - 2 (a - \frac{1}{a}) = (a - \frac{1}{a}) (a^{2} + 1 + \frac{1}{a^{2}} - 2) = (a - \frac{1}{a}) (a^{2} - 1 + \frac{1}{a^{2}})$

Question 27:

Factorize:
2a³ + 16b³ − 5a − 10b

Answer 27:

$2 a^{3} + 16 b^{3} - 5 a - 10 b = 2 [a^{3} + 8 b^{3}] - 5 (a + 2 b) = 2 [a^{3} + {(2 b)}^{3}] - 5 (a + 2 b) = 2 (a + 2 b) [a^{2} - a \times 2 b + {(2 b)}^{2}] - 5 (a + 2 b) = 2 (a + 2 b) (a^{2} - 2 a b + 4 b^{2}) - 5 (a + 2 b) = (a + 2 b) [2 (a^{2} - 2 a b + 4 b^{2}) - 5]$

Question 28:

Factorise
a⁶ + b⁶

Answer 28:

$a^{6} + b^{6} = {(a^{2})}^{3} + {(b^{2})}^{3} = (a^{2} + b^{2}) [{(a^{2})}^{2} - a^{2} b^{2} + {(b^{2})}^{2}] = (a^{2} + b^{2}) (a^{4} - a^{2} b^{2} + b^{4})$

Question 29:

Factorise
a¹² – b¹²

Answer 29:

a¹² – b^{12

$= (a^{6} + b^{6}) (a^{6} - b^{6}) = [{(a^{2})}^{3} + {(b^{2})}^{3}] [{(a^{3})}^{2} - {(b^{3})}^{2}] = [(a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})] [(a^{3} - b^{3}) (a^{3} + b^{3})] = [(a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})] [(a - b) (a^{2} + b^{2} + a b) (a + b) (a^{2} + b^{2} - a b)] = (a - b) (a^{2} + b^{2} + a b) (a + b) (a^{2} + b^{2} - a b) (a^{2} + b^{2}) (a^{4} + b^{4} - a^{2} b^{2})$}

Question 30:

Factorise
x⁶ – 7x³– 8

Answer 30:

Let $x^{3} = y$
So, the equation becomes
$y^{2} - 7 y - 8 = y^{2} - 8 y + y - 8 = y (y - 8) + (y - 8) = (y - 8) (y + 1) = (x^{3} - 8) (x^{3} + 1) = (x - 2) (x^{2} + 4 + 2 x) (x + 1) (x^{2} + 1 - x)$

Question 31:

Factorise
x³ – 3x²+ 3x + 7

Answer 31:

x³ – 3x²+ 3x + 7
$= x^{3} - 3 x^{2} + 3 x + 7 = x^{3} - 3 x^{2} + 3 x + 8 - 1 = x^{3} - 3 x^{2} + 3 x - 1 + 8 = (x^{3} - 3 x^{2} + 3 x - 1) + 8 = {(x - 1)}^{3} + 2^{3} = (x - 1 + 2) [{(x - 1)}^{2} + 4 - 2 (x - 1)] = (x + 1) [x^{2} + 1 - 2 x + 4 - 2 x + 2] = (x + 1) (x^{2} - 4 x + 7)$

Question 32:

Factorise
(x +1)³ + (x– 1)³

Answer 32:

(x +1)³ + (x– 1)^{3

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

Question 33:

Factorise
(2a +1)³ + (a– 1)³

Answer 33:

(2a +1)³ + (a– 1)^{3

$= (2 a + 1 + a - 1) [{(2 a + 1)}^{2} + {(a - 1)}^{2} - (2 a + 1) (a - 1)] = (3 a) [4 a^{2} + 1 + 4 a + a^{2} + 1 - 2 a - 2 a^{2} + 2 a - a + 1] = 3 a [3 a^{2} + 3 a + 3] = 9 a (a^{2} + a + 1)$}

Question 34:

Factorise
8(x +y)³ – 27(x– y)³

Answer 34:

8(x +y)³ – 27(x– y)^{3

$= {[2 (x + y)]}^{3} - {[3 (x - y)]}^{3} = (2 x + 2 y - 3 x + 3 y) [4 {(x + y)}^{2} + 9 {(x - y)}^{2} + 6 (x^{2} - y^{2})] = (- x + 5 y) [4 (x^{2} + y^{2} + 2 x y) + 9 (x^{2} + y^{2} - 2 x y) + 6 (x^{2} - y^{2})] = (- x + 5 y) [4 x^{2} + 4 y^{2} + 8 x y + 9 x^{2} + 9 y^{2} - 18 x y + 6 x^{2} - 6 y^{2}] = (- x + 5 y) (19 x^{2} + 7 y^{2} - 10 x y)$}

Question 35:

Factorise
(x +2)³ + (x– 2)³

Answer 35:

(x +2)³ + (x– 2)^{3

$= (x + 2 + x - 2) [{(x + 2)}^{2} + {(x - 2)}^{2} - (x^{2} - 4)] = 2 x (x^{2} + 4 + 4 x + x^{2} + 4 - 4 x - x^{2} + 4) = 2 x (x^{2} + 12)$}

Question 36:

Factorise
(x + 2)³ – (x– 2)³

Answer 36:

(x + 2)³ – (x– 2)^{3

$= (x + 2 - x + 2) [{(x + 2)}^{2} + {(x - 2)}^{2} + (x^{2} - 4)] = 4 [x^{2} + 4 + 4 x + x^{2} + 4 - 4 x + x^{2} - 4] = 4 (3 x^{2} + 4)$}

Question 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 37:

$LHS : \frac{0.85 \times 0.85 \times 0.85 + 0.15 \times 0.15 \times 0.15}{0.85 \times 0.85 - 0.85 \times 0.15 + 0.15 \times 0.15} = \frac{{(0.85)}^{3} + {(0.15)}^{3}}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} We know a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) Here a = 0.85, b = 0.15$
$\frac{{(0.85)}^{3} + {(0.15)}^{3}}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} = \frac{(0.85 + 0.15) ({(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2})}{{(0.85)}^{2} - 0.85 \times 0.15 + {(0.15)}^{2}} = 0.85 + 0.15 = 1 : RHS$
Thus, LHS = RHS

Question 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 38:

$\frac{59 \times 59 \times 59 - 9 \times 9 \times 9}{59 \times 59 + 59 \times 9 + 9 \times 9} = \frac{{(59)}^{3} - 9^{3}}{59^{2} + 59 \times 9 + 9^{2}}$
$We know a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) Here a = 59, b = 9 So, \frac{(59 - 9) (59^{2} + 9^{2} + 59 \times 9)}{59^{2} + 9^{2} + 59 \times 9} = 59 - 9 = 50 : RHS$
Thus, LHS=RHS

RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3F