Question:
If $f(x)=x^{3}-\frac{1}{x^{3}}$, show that $f(x)+f\left(\frac{1}{x}\right)=0$
Solution:
Given:
$f(x)=x^{3}-\frac{1}{x^{3}} \quad \ldots(\mathrm{i})$
Thus,
$f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}$
$=\frac{1}{x^{3}}-\frac{1}{\frac{1}{x^{3}}}$
$\therefore f\left(\frac{1}{x}\right)=\frac{1}{x^{3}}-x^{3} \ldots$ (ii)
$f(x)+f\left(\frac{1}{x}\right)=\left(x^{3}-\frac{1}{x^{3}}\right)+\left(\frac{1}{x^{3}}-x^{3}\right)$
$=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}=0$
Hence, $f(x)+f\left(\frac{1}{x}\right)=0$.
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