Question

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

Answer 1

Given: $x=1-\sqrt{2}$

$\Rightarrow \frac{1}{x}=\frac{1}{1-\sqrt{2}}$

On rationalizing, we get,

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

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

Hence,

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

Thus, $(x-\frac{1}{x}{)}^3=2^3=8$