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