Solve this

Question:

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

 

Solution:

Given: $f(x)=x^{3}-\frac{1}{x^{3}}$

Need to prove: $f(x)+f\left(\frac{1}{x}\right)=0$

Replacing $x$ by $\frac{1}{x}$ we get,

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

Now according to the problem,

$f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}$

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

 

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