Question

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

Answer 1

$f\left(x\right)=x^3-\frac{1}{x^3}\dots \dots \left(i\right)$
Now,
$f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^3-\frac{1}{{\left(\frac{1}{x}\right)}^3}=\frac{1}{x^3}-\frac{1}{\frac{1}{x^3}}\Rightarrow f\left(\frac{1}{x}\right)=\frac{1}{x^3}-x^3\dots \dots \left(ii\right)$
Adding equation (i) and equation (ii), we get
$f\left(x\right)+f\left(\frac{1}{x}\right)=(x^3-\frac{1}{x^3})+(\frac{1}{x^3}-x^3)=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3=0\therefore f\left(x\right)+f\left(\frac{1}{x}\right)=0$
Hence, proved.