Question

(1/x+y/3)3 expand using identity

Answer 1

Hey mate

here is your answer

$= {( \frac{1}{x} + \frac{y}{3} )}^{3} \\ = \frac{1}{ {x}^{3} } + \frac{ {y}^{3} }{27} + 3 \times \frac{1}{x} \times \frac{y}{3} ( \frac{1}{x} + \frac{y}{3} ) \\ = \frac{1}{ {x}^{3} } + \frac{ {y}^{3} }{27} + \frac{x}{y} ( \frac{1}{x} + \frac{y}{3} \: ) \\ = \frac{1}{ {x}^{3} } + \frac{ {y}^{3} }{27} + \frac{1}{y} + \frac{x}{3}$

hope it helps u .......

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.☺