Question

If $f\left(x\right)=x^3-\frac{1}{x^3},$ then $f\left(x\right)+f\left(\frac{1}{x}\right)$ is equal to

(a) 2x³

(b) $\frac{2}{x^3}$

(c) 0

(d) 1

Answer 1

$$\begin{gathered} f\left(x\right)=x^3-\frac{1}{x^3} \\ \Rightarrow f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^3-{\left(\frac{1}{{\displaystyle \frac{1}{x}}}\right)}^3 \\ f\left(\frac{1}{x}\right)=\frac{1}{x^3}-x^3 \\ \therefore f\left(x\right)+f\left(\frac{1}{x}\right)=x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3 \\ \mathrm{i}.\mathrm{e}f\left(x\right)+f\left(\frac{1}{x}\right)=0 \end{gathered}$$
Hence, the correct answer is option C.