Question

If $x-\frac{1}{x}=\sqrt{3}$, then $x^3-\frac{1}{x^3}$ equals

Answer 1

Given

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

on cubing both sides

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

$\Rightarrow x^3-3.x^2,\frac{1}{x}+3.x.\frac{1}{x^2}+\frac{1}{x^3}=3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}+3(x+\frac{1}{x})=3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}=3\sqrt{3}-3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}=0$