If f(x) =

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, $f(x)+f\left(\frac{1}{x}\right)=0$.