Question:
If $f(x)=x^{3}-\frac{1}{x^{3}}$ then show that $f(x)+f\left(\frac{1}{x}\right)=0$
Solution:
Given: $f(x)=x^{3}-\frac{1}{x^{3}}$
Need to prove: $f(x)+f\left(\frac{1}{x}\right)=0$
Replacing $x$ by $\frac{1}{x}$ we get,
$f\left(\frac{1}{x}\right)=\frac{1}{x^{3}}-\frac{1}{\frac{1}{x^{3}}}=\frac{1}{x^{3}}-x^{3}$
Now according to the problem,
$f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}$
$\Rightarrow f(x)+f\left(\frac{1}{x}\right)=0$ [Proved]