Question
Simplify the following: $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
✓
Answer
Given $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$
We shall use the identity $a^3 - b^3 = (a - b)(a^2 + b^2+ ab)$
Here $\text{a}=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big),\ \text{b=}\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)$ By applying identity we get $=\bigg(\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg)$
$\bigg[\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^2+\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg]$
$=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}-\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Bigg[\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2+\frac{2\text{xy}}{6}\Big)$
$+\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2-\frac{2\text{xy}}{6}\bigg)+\bigg(\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{y}}{3}\Big)^2\bigg)\Bigg]$
$=\frac{2\text{y}}{3}\Big[\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}\Big)+\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}\Big)+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
By rearranging the variable we get $=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{x}^2}{4}+\frac{\text{x}^2}{9}\Big]$
$=\frac{2\text{xy}}{3}\Big[\frac{3\text{x}^2}{4}+\frac{\text{y}^2}{9}\Big]$
$=\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}$ Hence the simplified value of $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$ is $\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}.$
We shall use the identity $a^3 - b^3 = (a - b)(a^2 + b^2+ ab)$
Here $\text{a}=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big),\ \text{b=}\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)$ By applying identity we get $=\bigg(\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg)$
$\bigg[\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^2+\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^2-\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)\bigg]$
$=\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}-\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)\Bigg[\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2+\frac{2\text{xy}}{6}\Big)$
$+\bigg(\Big(\frac{\text{x}}{2}\Big)^2+\Big(\frac{\text{y}}{3}\Big)^2-\frac{2\text{xy}}{6}\bigg)+\bigg(\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{y}}{3}\Big)^2\bigg)\Bigg]$
$=\frac{2\text{y}}{3}\Big[\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}\Big)+\Big(\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}\Big)+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
$=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}-\frac{2\text{xy}}{6}+\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}\Big]$
By rearranging the variable we get $=\frac{2\text{y}}{3}\Big[\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}+\frac{\text{x}^2}{4}+\frac{\text{x}^2}{9}\Big]$
$=\frac{2\text{xy}}{3}\Big[\frac{3\text{x}^2}{4}+\frac{\text{y}^2}{9}\Big]$
$=\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}$ Hence the simplified value of $\Big(\frac{\text{x}}{2}+\frac{\text{y}}{3}\Big)^3-\Big(\frac{\text{x}}{2}-\frac{\text{y}}{3}\Big)^3$ is $\frac{\text{x}^2\text{y}}{2}+\frac{2\text{y}^3}{27}.$
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