Question

If $x-\frac{1}{x}=3,$ then the value of $x^3-\frac{1}{x^3}$ is

Answer 1

Given $x-\frac{1}{x}=3$

Cubing on both sides

$(x-\frac{1}{x}{)}^3=27$

$⟹x^3-3\left(x\right)(\frac{1}{x}{)}^2+3(x{)}^2(\frac{1}{x})-(\frac{1}{x}{)}^3=27$

$⟹x^3-\frac{3}{x}+3x-\frac{1}{x^3}=27$

$⟹x^3-\frac{1}{x^3}=27-3(x-\frac{1}{x})=27-3\left(3\right)=27-9=18$