Question

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

Answer 1

$x=1-\sqrt{2}$

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

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

$=-\left(1+\sqrt{2}\right)$

$=-1-\sqrt{2}$

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

$=1-\sqrt{2}+1+\sqrt{2}$

$=2$

${\left(x-\frac{1}{x}\right)}^3={\left(2\right)}^3=8$