Question

Expand (1/x+y/3)^3 using the correct identity.

Answer 1

(1/x + y/3)³

[ (a+b)³ = a³+3a²b+3ab²+b³)]

→ (1/x)³+3(1/x)²(y/3)+3(1/x)(y/3)²+(y/3)³

→ 1/x³ + y/x² + y²/3x + y³/27
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Answer 2

Step-by-step explanation:

Solution:-

Given expression

$\rm \bigg( \frac{1}{x} + \frac{y}{3} \bigg) ^{3}$

∵ (a + b)³ = a³ + b³ + 3ab(a + b)

Where, we have to put in our expression a = 1/x and b = y/3.

$\rm\therefore \: \bigg( \frac{1}{x} + \frac{y}{3} \bigg)^{3}$

$\rm= \bigg( \frac{1}{x} \bigg) ^{3} + \bigg( \frac{y}{3} \bigg) ^{3} + 3 \times \frac{1}{x} \times \frac{y}{3} \bigg( \frac{1}{x} + \frac{y}{3} \bigg)^{3}$

$\rm = \frac{1}{ {x}^{3} } + \frac{ {y}^{2} }{27} + \frac{y}{x} \bigg( \frac{1}{x} + \frac{y}{3} \bigg)$

$\rm= \frac{1}{ {x}^{3} } + \frac{ {y}^{3} }{27} + \frac{y}{ {x}^{2} } + \frac{ {y}^{2} }{3x}$

$\rm= \frac{1}{ {x}^{3} } + \frac{y}{ {x}^{2} } + \frac{ {y}^{2} }{3x} + \frac{ {y}^{3} }{27} \: \bf{Ans.}$

Used formulae:-

(a + b)³ = x³ + y³ + 3xy(x + y).

I hope it's help you.☺