can you help?

I can't remember how to solve this problem, but I'm pretty sure the answer is 14.

Nevermind, I do remember how to solve this. 

First let's cube \(x-\frac{1}{x}=2\)

\((x-\frac{1}{x})^3=2^3\)

\(x^3-3x+\frac{3}{x}-\frac{1}{x^3}=8\)

\(x^3-\frac{1}{x^3}=8+3(x-\frac{1}{x})\)

\(x^3-\frac{1}{x^3}=8+3(2)\)

\(x^3-\frac{1}{x^3}=\boxed{14}\)