Question

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

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

Answer 1

The function is given as $f\left(x\right)=x+\frac{1}{x}$.

Simplifying the LHS of ${\left[f\left(x\right)\right]}^3=f\left(x^3\right)+3f\left(\frac{1}{x}\right)$.

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

$=x^3+\frac{1}{x^3}+3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}$

$=x^3+\frac{1}{x^3}+3x+3\times \frac{1}{x}$

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

$=f\left(x^3\right)+3f\left(x\right)$

$=RHS$