Question:
If $f(x)=x^{3}-\frac{1}{x^{3}}$, then $f(x)+f\left(\frac{1}{x}\right)$ is equal to
(a) $2 x^{3}$
(b) $\frac{2}{x^{3}}$
(c) 0
(d) 1
Solution:
$f(x)=x^{3}-\frac{1}{x^{3}}$
$\Rightarrow f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\left(\frac{1}{\frac{1}{x}}\right)^{3}$
$f\left(\frac{1}{x}\right)=\frac{1}{x^{3}}-x^{3}$
$\therefore f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}$
i. e $f(x)+f\left(\frac{1}{x}\right)=0$
Hence, the correct answer is option C.