If $(x - \frac {1}{x})^2 = 2$ and $x > 0$ what is $x^4 - \frac{1}{x^4}$? If $(x - \frac {1}{x})^2 = 2$ and $x &gt; 0$ what is $x^4 - \frac{1}{x^4}$?

I tried factorising the $x^4 - \frac{1}{x^4}$ expression and I got $(x^2 - \frac {1}{x^2})(x - \frac {1}{x})(x + \frac {1}{x})$ but I'm not sure what to do from here. 

Thanks!
$$x^2+\frac{1}{x^2}=4,$$ which gives
$$x^2=2+\sqrt3$$ or
$$x^2=2-\sqrt3.$$
Thus, $$x^4-\frac{1}{x^4}=4\left(x^2-\frac{1}{ x^2}\right)=8\sqrt3$$ or
$$x^4-\frac{1}{x^4}=-8\sqrt3.$$
Note that by expanding we have $2 = \left(x - \frac{1}{x} \right)^2 = x^2 - 2\frac{x}{x} + \frac{1}{x^2} = x^2 + \frac{1}{x^2} - 2$. So we have $x^2 + \frac{1}{x^2} = 4$, and squaring this again gives $x^4 + 2 + \frac{1}{x^4}=16$. So our answer is $14$.