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

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

To do:

We have to find the value of \( x^{3}- \frac{1}{x^{3}} \).

Solution:

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

Cubing both sides, we get,

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

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

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

$x^{3}-\frac{1}{x^{3}}-3 \times(3+2 \sqrt{2})=27+16 \sqrt{2}+54 \sqrt{2}+72$

$x^{3}+\frac{1}{x^{3}}-3(3+2 \sqrt{2})=99+70 \sqrt{2}$

$x^{3}+\frac{1}{x^{3}}=99+70 \sqrt{2}+9+6 \sqrt{2}$

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

The value of $x^{3}+\frac{1}{x^{3}}$ is $108+76 \sqrt{2}$.