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

Given, $x=1+\sqrt{2}$.

Now,

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

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

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

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

$=(1+\sqrt{2}-\sqrt{2}+1{)}^3=2^3=8$.

Therefore, option $C$ is correct.