Question

If $f\left(x\right)=x^3-\frac{1}{x^3}$, prove that $f\left(x\right)+f\left(\frac{1}{x}\right)=0$

Answer 1

$f\left(x\right)=x^2-\frac{1}{x^3}$

$\Rightarrow f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^3-\frac{1}{{\left(\frac{1}{x}\right)}^3}=\frac{1}{x^3}-\frac{1}{1/x^3}$

$=\frac{1}{x^3}-x^3$

$\therefore f\left(x\right)+f\left(\frac{1}{x}\right)=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3$

$\therefore f\left(x\right)+f\left(1/x\right)=0$