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 $x-\frac{1}{x}=2$

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

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

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

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

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