If $$x = 2 + \sqrt3$$ then the value of $$x^3 + x^{-3}$$ is:

Given, $$x=2+\sqrt{3}$$

$$=$$>  $$\ \frac{1}{x}=\ \frac{1}{2+\sqrt{3}}$$

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

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

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

$$\therefore\ x+\frac{1}{x}=(2+\sqrt{3})+(2-\sqrt{3})=4$$

$$=$$>  $$\left(\ x+\frac{1}{x}\right)^3=4^3$$

$$=$$>  $$x^3+\frac{\ 1}{x^3}+3.x.\frac{1}{x}\left(\ x+\frac{1}{x}\right)=64$$

$$=$$>  $$x^3+\frac{\ 1}{x^3}+3\left(4\right)=64$$

$$=$$>  $$x^3+x^{-3}=52$$

Hence, the correct answer is Option D