Question

If $x=3-2\sqrt{2}$, Find $(x^3-\frac{1}{x^3})$

Answer 1

$x=3-2\sqrt{2}$

$\frac{1}{x}=\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

$x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)$

Substituting the values of $x=3-2\sqrt{2}$ and $\frac{1}{x}=3+2\sqrt{2}$ we get

$=(3-2\sqrt{2}-3-2\sqrt{2})({(3-2\sqrt{2})}^2+{(3+2\sqrt{2})}^2+1)$

$=(-4\sqrt{2})(9+8-12\sqrt{2}+9+8+12\sqrt{2}+1)$

$=(-4\sqrt{2})\left(35\right)$

$=-140\sqrt{2}$