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,

\[f\left( \frac{1}{x} \right) = \left( \frac{1}{x} \right)^3 - \frac{1}{\left( \frac{1}{x} \right)^3}\] \[= \frac{1}{x^3} - \frac{1}{\frac{1}{x^3}}\]

\[\therefore f\left( \frac{1}{x} \right) = \frac{1}{x^3} - x^3\]  ...(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,

\[f\left( x \right) + f\left( \frac{1}{x} \right) = 0\]