Question

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

Answer 1

$x=3+2\sqrt{2}$ 1)

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

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

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

$x^2+\frac{1}{x^2}=(x-\frac{1}{x}{)}^2+2x\frac{1}{x}=(x-\frac{1}{x}{)}^2+2$ (4)

Substituting (4) in (3)

$x^3-\frac{1}{x^3}=(x-\frac{1}{x})\left(\right(x-\frac{1}{x}{)}^2+2+1)$ (5)

Substituting (1) and (2) in (5)

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

$x^3-\frac{1}{x^3}=\left(4\sqrt{2}\right)\left(35\right)=140\sqrt{2}$