Question

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

Answer 1

$$\begin{gathered} x=2-\sqrt{3}.....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2+\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2+\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{x}=2+\sqrt{3}.....\left(2\right) \end{gathered}$$
Subtracting (2) from (1), we get
$$\begin{gathered} x-\frac{1}{x}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right) \\ \Rightarrow x-\frac{1}{x}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3} \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^3={\left(-2\sqrt{3}\right)}^3=-24\sqrt{3} \end{gathered}$$
Thus, the value of ${\left(x-\frac{1}{x}\right)}^3$ is $-24\sqrt{3}$.