Question

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

Answer 1

$108+76\sqrt{2}$

Given , $x-\frac{1}{x}=3+2\sqrt{2}\dots \left(i\right)$,
Taking cube on both side,
${(x-\frac{1}{x})}^3={(3+2\sqrt{2})}^3$,
Using identity,
$(a-b{)}^3=a^3-b^3-3ab(a-b\left)\right(a+b{)}^3=a^3+b^3+3ab(a+b)\therefore {(x-\frac{1}{x})}^3=x^3-{\left(\frac{1}{x}\right)}^3-3\times x\times \frac{1}{x}(x-\frac{1}{x})$
$=3^3+(2\sqrt{2}{)}^3+3\times 3\times 2\sqrt{2}(3+2\sqrt{2})$
(from equation $1$)
$x^3-\frac{1}{x^3}-3(3+2\sqrt{2})=27+16\sqrt{2}+18\sqrt{2}+(3+2\sqrt{2})$
$x^3-\frac{1}{x^3}-9-6\sqrt{2}=27+16\sqrt{2}+54\sqrt{2}+72$
$x^3-\frac{1}{x^3}-9-6\sqrt{2}=99+70\sqrt{2}$
$x^3-\frac{1}{x^3}=99+9+70\sqrt{2}+6\sqrt{2}$
$x^3-\frac{1}{x^3}=108+76\sqrt{2}$