Question

Find the cube of
$(i) \: 3a + 5b \\ (ii) \: 3a + \frac{1}{3a} \\ (iii) \: 2x - \frac{1}{2x} + y$
(Ch- Expansion)​

Answer 1

(3a + 5b)³

= (3a)³ + (5b)³ + 3(3a)(5b) (3a+5b)

= 27a³ +125b³ +45ab(3a+5b)

= 27a³ +127b³ +135a²b+225ab²

(3a + 1/3a)³

= (3a)³+(1/3a)³ + 3(3a)(1/3a)(3a+1/3a)

= 27a³+1/27a³ +3(3a+1/3a)

= 27a³+ 1/27a³ + 9a + 3/3a

$= {27a}^{3} + \frac{1}{27} {a}^{3} + 9a + \frac{1}{a}$

Answer 2

Answer:

(i)27a³ + 125b³ + 135a²b + 225ab²

(ii)27a³ + 1/27a³ + 9a + 1/a

(iii)8x³ - 1/8x³ + 6x - 3/2x + y³+ 12x²y - 6y + 3y/4x + 6y²-3y²/2x

Step-by-step explanation:

(i)(3a+5b)³

= (3a)³ + (5b)³ + 3.3a.5b(3a+5b)               [As (a+b)³=a³ + b³ + 3ab(a+b)]

=27a³ + 125b³ + 135a²b + 225ab²

Ans:-27a³ + 125b³ + 135a²b + 225ab²

(ii)(3a + 1/3a)³

= (3a)³ + (1/3a)³ + 3.3a.1/3a(3a + 1/3a)       [As (a+b)³=a³ + b³ + 3ab(a+b)]

=27a³ + 1/27a³ + 9a + 1/a

Ans:-27a³ + 1/27a³ + 9a + 1/a

(iii) (2x - 1/2x + y)³

= (2x-1/2x)³ + y³ + 3.(2x-1/2x).y{(2x-1/2x)+y}        [As (a+b)³=a³ + b³ + 3ab(a+b)]

=m³ + y³ + 3.m.y(m+y)                    [Let m=(2x-1/2x)]

= m³ + y³ + 3m²y + 3my²

= (2x-1/2x)³ + y³ + 3(2x-1/2x)².y + 3(2x-1/2x).y²

=(2x)³ - (1/2x)³ + 3.2x.1/2x(2x - 1/2x) + y³ + 3(4x²-2+1/4x²).y + 6y²-3y²/2x              

=8x³ - 1/8x³ + 6x - 3/2x + y³+ 12x²y - 6y + 3y/4x + 6y²-3y²/2x

Ans:-8x³ - 1/8x³ + 6x - 3/2x + y³+ 12x²y - 6y + 3y/4x + 6y²-3y²/2x

(If you want to take common then take from these answers)