Question
Expand: $\left(\frac{x}{3}-\frac{3}{x}\right)^3$

Answer

Here, $\mathrm{a}=\frac{x}{3}$ and $\mathrm{b}=\frac{3}{x}$
$\begin{aligned}
\left(\frac{x}{3}-\frac{3}{x}\right)^3= & \left(\frac{x}{3}\right)^3-3\left(\frac{x}{3}\right)^2\left(\frac{3}{x}\right) \\
& +3\left(\frac{x}{3}\right)\left(\frac{3}{x}\right)^2-\left(\frac{3}{x}\right)^3 \\
& \quad \ldots\left[(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right] \\
& \frac{x^3}{27}-3\left(\frac{x}{3}\right)\left(\frac{x}{3}\right)\left(\frac{3}{x}\right)+3\left(\frac{x}{3}\right)\left(\frac{3}{x}\right)\left(\frac{3}{x}\right)-\frac{27}{x^3} \\
= & \frac{x^3}{27}-3\left(\frac{x}{3}\right)+3\left(\frac{3}{x}\right)-\frac{27}{x^3} \\
= & \frac{x^3}{27}-x+\frac{9}{x}-\frac{27}{x^3}
\end{aligned}$

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