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

EXERCISE 3G

PAGE NO-136

Question 1:

Find the product:
(x + y − z) (x² + y² + z² − xy + yz + zx)

Answer 1:

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

Question 2:

Find the product:
(x – y − z) (x² + y² + z² + xy – yz + xz)

Answer 2:

(x – y − z) (x² + y² + z² + xy – yz + xz)
$= (x + (- y) + (- z)) (x^{2} + y^{2} + z^{2} + x y - y z + x z) We know (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c Here, a = x, b = - y, c = - z (x + (- y) + (- z)) (x^{2} + y^{2} + z^{2} + x y - y z + x z) = x^{3} - y^{3} - z^{3} - 3 x y z$

Question 3:

Find the product:
(x − 2y + 3) (x² + 4y² + 2xy − 3x + 6y + 9)

Answer 3:

$(x − 2y + 3) (x^{2} + 4 y^{2} + 2 x y - 3 x + 6 y + 9) = (x − 2y + 3) (x^{2} + 4 y^{2} + 9 + 2 x y + 6 y - 3 x) = [x + (- 2 y) + 3] [x^{2} + {(- 2 y)}^{2} + {(3)}^{2} - x \times (- 2 y) - (- 2 y) \times 3 - 3 \times x] = {(x)}^{3} + {(- 2 y)}^{3} + {(3)}^{3} - 3 (x) (- 2 y) (3) = x^{3} - 8 y^{3} + 27 + 18 x y$

Question 4:

Find the product:
(3x – 5y + 4) (9x² + 25y² + 15xy − 20y + 12x + 16)

Answer 4:

$(3 x - 5 y + 4) (9 x^{2} + 25 y^{2} + 15 x y - 20 y + 12 x + 16) = (3 x + (- 5 y) + 4) (9 x^{2} + 25 y^{2} + 16 + 15 x y - 20 y + 12 x)$
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = a^{3} + b^{3} + c^{3} - 3 a b c Here, a = 3 x, b = - 5 y, c = 4$
$(3 x + (- 5 y) + 4) (9 x^{2} + 25 y^{2} + 16 + 15 x y - 20 y + 12 x) = {(3 x)}^{3} + {(- 5 y)}^{3} + 4^{3} - 3 \times 3 x (- 5 y) (4) = 27 x^{3} - 125 y^{3} + 64 + 180 x y$

Question 5:

Factorize:
125a³ + b³ + 64c³ − 60abc

Answer 5:

$125 a^{3} + b^{3} + 64 c^{3} - 60 a b c = {(5 a)}^{3} + {(b)}^{3} + {(4 c)}^{3} - 3 \times 5 a \times b \times 4 c = (5 a + b + 4 c) [{(5 a)}^{2} + {(b)}^{2} + {(4 c)}^{2} - 5 a \times b - b \times 4 c - 5 a \times 4 c] = (5 a + b + 4 c) (25 a^{2} + b^{2} + 16 c^{2} - 5 a b - 4 b c - 20 a c)$

Question 6:

Factorize:
a³ + 8b³ + 64c³ − 24abc

Answer 6:

$a^{3} + 8 b^{3} + 64 c^{3} - 24 a b c = a^{3} + {(2 b)}^{3} + {(4 c)}^{3} - 3 \times a \times 2 b \times 4 c = (a + 2 b + 4 c) [a^{2} + {(2 b)}^{2} + {(4 c)}^{2} - a \times 2 b - 2 b \times 4 c - 4 c \times a] = (a + 2 b + 4 c) (a^{2} + 4 b^{2} + 16 c^{2} - 2 a b - 8 b c - 4 c a)$

Question 7:

Factorize:
1 + b³ + 8c³ − 6bc

Answer 7:

$1 + b^{3} + 8 c^{3} - 6 b c = {(1)}^{3} + {(b)}^{3} + {(2 c)}^{3} - 3 \times 1 \times b \times 2 c = (1 + b + 2 c) [1^{2} + b^{2} + {(2 c)}^{2} - 1 \times b - b \times 2 c - 1 \times 2 c] = (1 + b + 2 c) (1 + b^{2} + 4 c^{2} - b - 2 b c - 2 c)$

Question 8:

Factorize:
216 + 27b³ + 8c³ − 108abc

Answer 8:

$216 + 27 b^{3} + 8 c^{3} - 108 a b c = {(6)}^{3} + {(3 b)}^{3} + {(2 c)}^{3} - 3 \times 6 \times 3 b \times 2 c = (6 + 3 b + 2 c) [6^{2} + {(3 b)}^{2} + {(2 c)}^{2} - 6 \times 3 b - 3 b \times 2 c - 2 c \times 6] = (6 + 3 b + 2 c) (36 + 9 b^{2} + 4 c^{2} - 18 b - 6 b c - 12 c)$

Question 9:

Factorize:
27a³ − b³ + 8c³ + 18abc

Answer 9:

$27 a^{3} - b^{3} + 8 c^{3} + 18 a b c = {(3 a)}^{3} + {(- b)}^{3} + {(2 c)}^{3} - 3 \times (3 a) \times (- b) \times (2 c) = [3 a + (- b) + 2 c] [{(3 a)}^{2} + {(- b)}^{2} + {(2 c)}^{2} - 3 a (- b) - (- b) 2 c - 3 a \times 2 c] = (3 a - b + 2 c) (9 a^{2} + b^{2} + 4 c^{2} + 3 a b + 2 b c - 6 a c)$

Question 10:

Factorize:
8a³ + 125b³ − 64c³ + 120abc

Answer 10:

$8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c = {(2 a)}^{3} + {(5 b)}^{3} + {(- 4 c)}^{3} - 3 \times (2 a) \times (5 b) \times (- 4 c) = (2 a + 5 b - 4 c) [{(2 a)}^{2} + {(5 b)}^{2} + {(- 4 c)}^{2} - (2 a) (5 b) - (5 b) (- 4 c) - (2 a) \times (- 4 c)] = (2 a + 5 b - 4 c) (4 a^{2} + 25 b^{2} + 16 c^{2} - 10 a b + 20 b c + 8 a c)$

Question 11:

Factorize:
8 − 27b³ − 343c³ − 126bc

Answer 11:

$8 - 27 b^{3} - 343 c^{3} - 126 b c = {(2)}^{3} + {(- 3 b)}^{3} + {(- 7 c)}^{3} - 3 \times (2) \times (- 3 b) \times (- 7 c) = [2 + (- 3 b) + (- 7 c)] [{(2)}^{2} + {(- 3 b)}^{2} + {(- 7 c)}^{2} - (2) (- 3 b) - (- 3 b) (- 7 c) - (2) (- 7 c)] = (2 - 3 b - 7 c) (4 + 9 b^{2} + 49 c^{2} + 6 b - 21 b c + 14 c)$

Question 12:

Factorize:
125 − 8x³ − 27y³ − 90xy

Answer 12:

$125 - 8 x^{3} - 27 y^{3} - 90 x y = 5^{3} + {(- 2 x)}^{3} + {(- 3 y)}^{3} - 3 \times 5 \times (- 2 x) \times (- 3 y) = [5 + (- 2 x) + (- 3 y)] [5^{2} + {(- 2 x)}^{2} + {(- 3 y)}^{2} - 5 \times (- 2 x) - (- 2 x) (- 3 y) - 5 \times (- 3 y)] = (5 - 2 x - 3 y) (25 + 4 x^{2} + 9 y^{2} + 10 x - 6 x y + 15 y)$

PAGE NO-137

Question 13:

Factorize:
$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c$

Answer 13:

$2 \sqrt{2} a^{3} + 16 \sqrt{2} b^{3} + c^{3} - 12 a b c = {(\sqrt{2} a)}^{3} + {(2 \sqrt{2} b)}^{3} + c^{3} - 3 \times (\sqrt{2} a) \times (2 \sqrt{2} b) \times (c) = (\sqrt{2} a + 2 \sqrt{2} b + c) [{(\sqrt{2} a)}^{2} + {(2 \sqrt{2} b)}^{2} + c^{2} - (\sqrt{2} a) \times (2 \sqrt{2} b) - (2 \sqrt{2} b) \times (c) - (\sqrt{2} a) \times (c)] = (\sqrt{2} a + 2 \sqrt{2} b + c) (2 a^{2} + 8 b^{2} + c^{2} - 4 a b - 2 \sqrt{2} b c - \sqrt{2} a c)$

Question 14:

Factorise:
27x³ – y³ – z³ – 9xyz

Answer 14:

$27 x^{3} - y^{3} - z^{3} - 9 x y z = {(3 x)}^{3} - y^{3} - z^{3} - 3 \times (3 x) \times (- y) \times (- z) W e k n o w, a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) a = 3 x, b = - y, c = - z {(3 x)}^{3} - y^{3} - z^{3} - 3 \times (3 x) \times (- y) \times (- z) = (3 x - y - z) (9 x^{2} + y^{2} + z^{2} + 3 x y - y z + 3 x z)$

Question 15:

Factorise:
$2 \sqrt{2} a^{3} + 3 \sqrt{3} b^{3} + c^{3} - 3 \sqrt{6} a b c$

Answer 15:

$2 \sqrt{2} a^{3} + 3 \sqrt{3} b^{3} + c^{3} - 3 \sqrt{6} a b c = {(\sqrt{2} a)}^{3} + {(\sqrt{3} b)}^{3} + c^{3} - 3 (\sqrt{2} a) (\sqrt{3} b) c We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x = \sqrt{2} a, y = \sqrt{3} b, z = c {(\sqrt{2} a)}^{3} + {(\sqrt{3} b)}^{3} + c^{3} - 3 (\sqrt{2} a) (\sqrt{3} b) c = (\sqrt{2} a + \sqrt{3} b + c) (2 a^{2} + 3 b^{2} + c^{2} - \sqrt{6 a b} - \sqrt{3 b c} - \sqrt{2} a c)$

Question 16:

Factorise:
$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c$

Answer 16:

$3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c = {(\sqrt{3} a)}^{3} + {(- b)}^{3} + {(- \sqrt{5} c)}^{3} - 3 (\sqrt{3} a) (- b) (- \sqrt{5} c) W e k n o w x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) H e r e, x = (\sqrt{3} a), y = (- b), z = (- \sqrt{5} c) 3 \sqrt{3} a^{3} - b^{3} - 5 \sqrt{5} c^{3} - 3 \sqrt{15} a b c = {(\sqrt{3} a)}^{3} + {(- b)}^{3} + {(- \sqrt{5} c)}^{3} - 3 (\sqrt{3} a) (- b) (- \sqrt{5} c) = (\sqrt{3} a - b - \sqrt{5} c) (3 a^{2} + b^{2} + 5 c^{2} + \sqrt{3} a b - \sqrt{5} b c + \sqrt{15} c)$

Question 17:

Factorize:
(a − b)³ + (b − c)³ + (c − a)³

Answer 17:

${(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}$
$Putting (a - b) = x, (b - c) = y and (c - a) = z, we get : {(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3} = x^{3} + y^{3} + z^{3} [Where (x + y + z) = (a - b) + (b - c) + (c - a) = 0] = 3 x y z [(x + y + z) = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z] = 3 (a - b) (b - c) (c - a)$

Question 18:

Factorise:
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3}$

Answer 18:

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z H e r e, x = (a - 3 b), y = (3 b - c), z = (c - a)$
${(a - 3 b)}^{3} + {(3 b - c)}^{3} + {(c - a)}^{3} = (a - 3 b + 3 b - c + c - a) [{(a - 3 b)}^{2} + {(3 b - c)}^{2} + {(c - a)}^{2} - (a - 3 b) (3 b - c) - (3 b - c) (c - a) - (c - a) (a - 3 b)] + 3 (a - 3 b) (3 b - c) (c - a) = 0 + 3 (a - 3 b) (3 b - c) (c - a) = 3 (a - 3 b) (3 b - c) (c - a)$

Question 19:

Factorize:
(3a − 2b)³ + (2b − 5c)³ + (5c − 3a)³

Answer 19:

$Put (3 a - 2 b) = x, (2 b - 5 c) = y and (5 c - 3 a) = z . W e have : x + y + z = 3 a - 2 b + 2 b - 5 c + 5 c - 3 a = 0 Now, {(3 a - 2 b)}^{3} + {(2 b - 5 c)}^{3} + {(5 c - 3 a)}^{3} = x^{3} + y^{3} + z^{3} = 3 x y z [Here, x + y + z = 0 . So, x^{3} + y^{3} + z^{3} = 3 x y z] = 3 (3 a - 2 b) (2 b - 5 c) (5 c - 3 a)$

Question 20:

Factorize:
(5a − 7b)³ + (9c − 5a)³ + (7b − 9c)³

Answer 20:

$Put (5 a - 7 b) = x, (9 c - 5 a) = z and (7 b - 9 c) = y . Here, x + y + z = 5 a - 7 b + 9 c - 5 a + 7 b - 9 c = 0 Thus, we have : {(5 a - 7 b)}^{3} + {(9 c - 5 a)}^{3} + {(7 b - 9 c)}^{3} = x^{3} + z^{3} + y^{3} = 3 x z y [When x + y + z = 0, x^{3} + y^{3} + z^{3} = 3 x y z .] = 3 (5 a - 7 b) (9 c - 5 a) (7 b - 9 c)$

Question 21:

Factorize:
a³(b − c)³ + b³(c − a)³ + c³(a − b)³

Answer 21:

$We have : a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} = {[a (b - c)]}^{3} + {[b (c - a)]}^{3} + {[c (a - b)]}^{3} Put a (b - c) = x b (c - a) = y c (a - b) = z Here, x + y + z = a (b - c) + b (c - a) + c (a - b) = a b - a c + b c - a b + a c - b c = 0 Thus, we have : a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} = x^{3} + y^{3} + z^{3} = 3 x y z [When x + y + z = 0, x^{3} + y^{3} + z^{3} = 3 x y z .] = 3 a (b - c) b (c - a) c (a - b) = 3 a b c (a - b) (b - c) (c - a)$

Question 22:

Evaluate
(i) (–12)³+ 7³+ 5³
(ii) (28)³ + (–15)³ + (–13)³

Answer 22:

(i) (–12)³+ 7³+ 5³
$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 12), y = 7, z = 5$
${(- 12)}^{3} + 7^{3} + 5^{3} = (- 12 + 7 + 5) [{(- 12)}^{2} + 7^{2} + 5^{2} - 7 (- 12) - 35 + 60] + 3 (- 12) \times 35 = 0 - 1260 = - 1260$

(ii) (28)³ + (–15)³ + (–13)³

$We know x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) + 3 x y z Here, x = (- 28), y = - 15, z = - 13$
${(28)}^{3} + {(- 15)}^{3} + {(- 13)}^{3} = (28 - 15 - 13) [{(28)}^{2} + {(- 15)}^{2} + {(- 13)}^{2} - 28 (- 15) - (- 15) (- 13) - 28 (- 13)] + 3 \times 28 (- 15) (- 13) = 0 + 16380 = 16380$

Question 23:

Prove that ${(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)$

Answer 23:

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

Question 24:

If a, b, c are all nonzero and a + b + c = 0, prove that $\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3$ .

Answer 24:

$a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$

Thus, we have:
$(\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b}) = \frac{a^{3} + b^{3} + c^{3}}{a b c}$
$= \frac{3 a b c}{a b c} = 3$

Question 25:

If a + b + c = 9 and a² + b² + c² = 35, find the value of (a³+ b³ + c³ – 3abc).

Answer 25:

a + b + c = 9
$\Rightarrow {(a + b + c)}^{2} = 9^{2} = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 81 \Rightarrow 35 + 2 (a b + b c + c a) = 81 \Rightarrow (a b + b c + c a) = 23$
We know,
(a³+ b³ + c³ – 3abc) = $(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
$= (9) (35 - 23) = 108$

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