Simplify each of the following:$(\frac{x}{2}+\frac{y}{3})^{3}-(\frac{x}{2}-\frac{y}{3})^{3}$
Given:
$(\frac{x}{2}+\frac{y}{3})^{3}-(\frac{x}{2}-\frac{y}{3})^{3}$
To do:
We have to simplify the given expression.
Solution:
$(\frac{x}{2}+\frac{y}{3})^{3}-(\frac{x}{2}-\frac{y}{3})^{3}=(\frac{x}{2})^{3}+(\frac{y}{3})^{3}+3 \times(\frac{x}{2})^{2} \times \frac{y}{3}+3\times(\frac{x}{2})\times(\frac{y}{3})^{2}-[(\frac{x}{2})^{3}-(\frac{y}{3})^{3}-3\times(\frac{x}{2})^{2}\times \frac{y}{3}+3\times(\frac{x}{2})\times(\frac{y}{3})^{2}]$
$=[\frac{x^{3}}{8}+\frac{y^{3}}{27}+3 \times \frac{x^{2}}{4} \times \frac{y}{3}+3 \times \frac{x}{2} \times \frac{y^{2}}{9}]-[\frac{x^{3}}{8}-\frac{y^{3}}{27}-3 \times \frac{x^{2}}{4} \times \frac{y}{3}+3 \frac{x}{2} \times \frac{y^{2}}{9}]$
$=\frac{x^{3}}{8}+\frac{y^{3}}{27}+\frac{x^{2} y}{4}+\frac{x y^{2}}{6}-\frac{x^{3}}{8}+\frac{y^{3}}{27}+\frac{x^{2} y}{4}-\frac{x y^{2}}{6}$
$=\frac{2 y^{3}}{27}+\frac{x^{2} y}{2}$
Hence, $(\frac{x}{2}+\frac{y}{3})^{3}-(\frac{x}{2}-\frac{y}{3})^{3}=\frac{2 y^{3}}{27}+\frac{x^{2} y}{2}$.
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