Question

If $f\left(x\right)=x+\frac{1}{x}$

prove that $\left[f\right(x){]}^3=f\left(x^3\right)+3f\left(\frac{1}{x}\right)$

Answer 1

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

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

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

To prove : $\left[f\right(x){]}^3=f(x^3)+3f\left(\frac{1}{x}\right)$

LHS

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

$=x^3+\frac{1}{x^3}+3\left[\frac{x^2}{x}\right]+3\left[\frac{x}{x^2}\right]$

$=x^3+\frac{1}{x^3}+3[x+\frac{1}{x}]$

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

LHS = RHS

Hence verified