# If f(x)=

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.