Question

( 1/x + y/3 )3 expand

Answer 1

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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).

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