Find the cube of each of the following binomial expressions:
$ \frac{1}{x}+\frac{y}{3} $

Given:

\( \frac{1}{x}+\frac{y}{3} \)

To do:

We have to find the cube of the given binomial expression.

Solution:

We know that,

$(a+b)^3=a^3 + b^3+ 3a^2b + 3ab^2$

Therefore,

$(\frac{1}{x}+\frac{y}{3})^{3}=(\frac{1}{x})^{3}+(\frac{y}{3})^{3}+3\times(\frac{1}{x})^{2} \times \frac{y}{3}+3 \times \frac{1}{x} \times (\frac{y}{3})^{2}$

$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+3 \times \frac{1}{x^{2}} \times \frac{y}{3}+3 \times \frac{1}{x} \times \frac{y^{2}}{9}$

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

Hence, $(\frac{1}{x}+\frac{y}{3})^{3}=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}$. 

Tutorialspoint

Updated on: 10-Oct-2022

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