Question
Expand : $\left(x+\frac{1}{x}\right)^3$
✓
Answer
Here, $a=x$ and $b=\frac{1}{x}$
$
\begin{aligned}
& \left(x+\frac{1}{x}\right)^3 \\
& =(x)^3+3(x)^2\left(\frac{1}{x}\right)+3(x)\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3 \\
& \quad \ldots\left[\because(\mathrm{a}+\mathrm{b})^3=\mathrm{a}^3+3 \mathrm{a}^2 \mathrm{~b}+3 \mathrm{ab}^2+\mathrm{b}^3\right] \\
& =x^3+3 x+3 x\left(\frac{1}{x^2}\right)+\frac{1}{x^3} \\
& =x^3+3 x+\frac{3}{x}+\frac{1}{x^3}
\end{aligned}
$
$
\begin{aligned}
& \left(x+\frac{1}{x}\right)^3 \\
& =(x)^3+3(x)^2\left(\frac{1}{x}\right)+3(x)\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3 \\
& \quad \ldots\left[\because(\mathrm{a}+\mathrm{b})^3=\mathrm{a}^3+3 \mathrm{a}^2 \mathrm{~b}+3 \mathrm{ab}^2+\mathrm{b}^3\right] \\
& =x^3+3 x+3 x\left(\frac{1}{x^2}\right)+\frac{1}{x^3} \\
& =x^3+3 x+\frac{3}{x}+\frac{1}{x^3}
\end{aligned}
$
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