Question

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

Answer 1

We have, $x+\frac{1}{x}=4$
Cubing both the sides, we get
${(x+\frac{1}{x})}^3=(4{)}^3$
$\Rightarrow x^3+\frac{1}{x^3}+3.x.\frac{1}{x}(x+\frac{1}{x})=64$
$\Rightarrow x^3+\frac{1}{x^3}+3\left(4\right)=64$
$\Rightarrow x^3+\frac{1}{x^3}=64-12$
$\Rightarrow x^3+\frac{1}{x^3}=52$