If $$x - \frac{1}{x} = \sqrt{77}$$, then one of the values of $$x^3 + \frac{1}{x^3}$$ is:

$$x - \frac{1}{x} = \sqrt{77}$$

$$\left(x-\frac{1}{x}\right)^2=77$$

$$x^2+\frac{1}{x^2}-2=77$$

$$x^2+\frac{1}{x^2}=79$$

$$x^2+\frac{1}{x^2}+2=81$$

$$\left(x+\frac{1}{x}\right)^2=81$$

$$x+\frac{1}{x}=9$$ or $$x+\frac{1}{x}=-9$$

When $$x+\frac{1}{x}=-9$$

$$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=-729$$

$$x^3+\frac{1}{x^3}+3\left(-9\right)=-729$$

$$x^3+\frac{1}{x^3}-27=-729$$

$$x^3+\frac{1}{x^3}=-702$$

Hence, the correct answer is Option B