Question
Simplify the following: $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$

Answer

Given $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$ We shall use the identity $a^3 + b^3 = (a - b)(a^2 + b^2- ab)$ Here $\text{a}=\Big(\text{x}+\frac{2}{\text{x}}\Big),\ \text{b}=\Big(\text{x}-\frac{2}{\text{x}}\Big)$ By applying identity we get $​​=\Big(\text{x}+\frac{2}{\text{x}}+\text{x}-\frac{2}{\text{x}}\Big)\Bigg[\Big(\text{x}+\frac{2}{\text{x}}\Big)^2+\Big(\text{x}-\frac{2}{\text{x}}\Big)^2$
$-\bigg(\Big(\text{x}+\frac{2}{\text{x}}\Big)\times\Big(\text{x}-\frac{2}{\text{x}}\Big)\bigg)\Bigg]$
$=\Big(\text{x}+\frac{2}{\text{x}}+\text{x}-\frac{2}{\text{x}}\Big)\bigg[\Big(\text{x}\times\text{x}+\frac{2}{\text{x}}\times\frac{2}{\text{x}}+2\times\text{x}\times\frac{2}{\text{x}}\Big)$
$+\Big(\text{x}\times\text{x}+\frac{2}{\text{x}}\times\frac{2}{\text{x}}-2\times\text{x}\times\frac{2}{\text{x}}\Big)-\Big(\text{x}^2+\frac{4}{\text{x}^2}\Big)\bigg]$
$=(2\text{x})\bigg[\Big(\text{x}^2+\frac{4}{\text{x}^2}+\frac{4\text{x}}{\text{x}}\Big)+\Big(\text{x}^2+\frac{4}{\text{x}^2}-\frac{4\text{x}}{\text{x}}\Big)-\Big(\text{x}^2-\frac{4}{\text{x}^2}\Big)\bigg]$
$=(2\text{x})\bigg[\text{x}^2+\frac{4}{\text{x}^2}+\frac{4\text{x}}{\text{x}}+\text{x}^2+\frac{4}{\text{x}^2}-\frac{4\text{x}}{\text{x}}-\text{x}^2+\frac{4}{\text{x}^2}\bigg]$ By rearranging the variable we get, $=(2\text{x})\Big[\text{x}^2+\frac{4}{\text{x}^2}+\frac{4}{\text{x}^2}+\frac{4}{\text{x}}^2\Big]$
$=2\text{x}\times\Big[\text{x}^2+\frac{12}{\text{x}^2}\Big]$
$=2\text{x}^2+\frac{24}{\text{x}}$ Hence the simplified value of $\Big(\text{x}+\frac{2}{\text{x}}\Big)^3+\Big(\text{x}-\frac{2}{\text{x}}\Big)^3$ is $2\text{x}^2+\frac{24}{\text{x}}.$

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