Question

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

Answer 1

Given:
$f\left(x\right)=x^3-\frac{1}{x^3}$ ...(i)
Thus,
$f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^3-\frac{1}{{\left(\frac{1}{x}\right)}^3}$

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

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

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

$=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3=0$

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