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Question:

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

 

 

Solution:

$x=2-\sqrt{3} \quad \ldots .(1)$

$\Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}}$

$\Rightarrow \frac{1}{x}=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

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

$\Rightarrow \frac{1}{x}=\frac{2+\sqrt{3}}{4-3}$

$\Rightarrow \frac{1}{x}=2+\sqrt{3} \quad \ldots(2)$

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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