find the value of

$x=2+\sqrt{3}$    ..........(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