The value of

Given,

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

Cubing $x-\frac{1}{x}=3+2 \sqrt{2}$ on both sides

We know that, $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

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

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

$x^{3}-\frac{1}{x^{3}}-3(3+2 \sqrt{2})=27+16 \sqrt{2}+18 \sqrt{2}(3+2 \sqrt{2})$

$x^{3}-\frac{1}{x^{3}}=27+16 \sqrt{2}+54 \sqrt{2}+72+9+6 \sqrt{2}$

$x^{3}-\frac{1}{x^{3}}=108+76 \sqrt{2}$

hence, the value of $x^{3}-\frac{1}{x^{3}}=108+76 \sqrt{2}$