Question

If $x-\frac{1}{x}=3+2\sqrt{2}$, find the value of $x^3-\frac{1}{x^3}$.

Answer 1

$x-\frac{1}{x}=3+2\sqrt{2}$

Cubing both sides,

${(x-\frac{1}{x})}^3=(3+2\sqrt{2}{)}^3$
$\Rightarrow x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=(3{)}^3+(2\sqrt{2}{)}^3+3\times 3\times 2\sqrt{2}(3+2\sqrt{2})$
$\Rightarrow x^3-\frac{1}{x^3}-3\times (3+2\sqrt{2})=27+16\sqrt{2}+18\sqrt{2}(3+2\sqrt{2})$
$\Rightarrow x^3-\frac{1}{x^3}-3\times (3+2\sqrt{2})=27+16\sqrt{2}+54\sqrt{2}+72$
$\Rightarrow x^3+\frac{1}{x^3}-3(3+2\sqrt{2})=99+70\sqrt{2}$
$\Rightarrow x^3+\frac{1}{x^3}=99+70\sqrt{2}+3(3+2\sqrt{2})$
$=99+70\sqrt{2}+9+6\sqrt{2}$
$=108+76\sqrt{2}$