If $x=2+\sqrt{3}$, find the value of $x^{3}+\frac{1}{x^{3}}$.
$x=2+\sqrt{3}$ ..........(1)
$\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}$
$\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$\Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^{2}}$
$\Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}$
$\Rightarrow \frac{1}{x}=2-\sqrt{3}$ ....(2)
Adding (1) and (2), we get
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4 \quad \ldots(3)$
Cubing both sides, we get
$\left(x+\frac{1}{x}\right)^{3}=4^{3}$
$\Rightarrow x^{3}+\frac{1}{x^{3}}+3 \times x \times \frac{1}{x}\left(x+\frac{1}{x}\right)=64$
$\Rightarrow x^{3}+\frac{1}{x^{3}}+3 \times 4=64 \quad$ [Using (3)]
$\Rightarrow x^{3}+\frac{1}{x^{3}}=64-12=52$
Thus, the value of $x^{3}+\frac{1}{x^{3}}$ is 52
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