If $$x = 2 + \sqrt5$$ then the value of $$(x^3 - x^{-3})$$ is:

Given, $$x=2+\sqrt{5}$$

$$=$$> $$\ \frac{1}{x}=\ \frac{1}{2+\sqrt{5}}$$

$$=$$> $$\ \frac{1}{x}=\ \frac{1}{2+\sqrt{5}}\times\ \frac{2-\sqrt{5}}{2-\sqrt{5}}$$

$$=$$> $$\ \frac{1}{x}=\ \frac{2-\sqrt{5}}{4-5}$$

$$=$$> $$\ \frac{1}{x}= \sqrt{5}-2$$

$$\therefore\ x-\frac{1}{x}=(2+\sqrt{5})-(\sqrt{5}-2)=4$$

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

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

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

$$=$$> $$x^3+x^{-3}=76$$

Hence, the correct answer is Option B