Question

If $x+\frac{1}{x}=4$, find the value of $x^3+\frac{1}{x^3}$.

Answer 1

Given, $x+\frac{1}{x}=4$ and $x^3+\frac{1}{x^3}=?$

On cubing both sides,

$x^3+\frac{1}{x^3}+3\left(x^2\right)\frac{1}{x}+3x\times \frac{1}{x^2}=64$

$⟹$ $x^3+\frac{1}{x^3}+3x+\frac{3}{x}=64$

$⟹$ $x^3+\frac{1}{x^3}=64-3(x+\frac{1}{x})$

$⟹$ $x^3+\frac{1}{x^3}=64-3\left(4\right)$

$⟹$ $x^3+\frac{1}{x^3}=64-12$

$⟹$ $x^3+\frac{1}{x^3}=52$.