Math, asked by Rips2400, 11 months ago

expand (1/x + y/3)³​

Answers

Answered by kuruish37
11

(1/x+y/3)^3

=(1/x)^3+(y/3)^3+3(1/x)^2(y/3)+3(1/x)(y/3)^2

=1/x^3+y^3/27+3y/3x^2+3y^2/9x

Answered by Salmonpanna2022
1

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.☺
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