Question
If $\text{x}=2-\sqrt{3},$ find the value of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^3.$
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Answer
$\text{x}=2-\sqrt{3}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}$
$=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{4-3}$
$=2-\sqrt{3}-2-\sqrt{3}$
$=-2\sqrt{3}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\big(-2\sqrt{3}\big)^3=-24\sqrt{3}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}$
$=2-\sqrt{3}-\frac{1}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=2-\sqrt{3}-\frac{2+\sqrt{3}}{4-3}$
$=2-\sqrt{3}-2-\sqrt{3}$
$=-2\sqrt{3}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\big(-2\sqrt{3}\big)^3=-24\sqrt{3}$
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