Show that $\dfrac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =2^\frac{1}{6}$ This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions.
Now I know that these two are equal, but I just can't prove it. I tried writing the LHS in terms of powers of $2$, but can't get any further to the desired result.

${\sqrt{8-4\sqrt3}\over\sqrt[3]{-20+12\sqrt3}}$
=${\left(8-4\sqrt3\right)^{1\over2}\over\left(-20+12\sqrt3\right)^{1\over3}}$
=${\left(8-4\sqrt3\right)^{3\over6}\over\left(-20+12\sqrt3\right)^{2\over6}}$
=$\left(\left(8-4\sqrt3\right)^3\over\left(-20+12\sqrt3\right)^2\right)^\frac16$
=$\sqrt[6]{\left(8-4\sqrt3\right)^3\over\left(-20+12\sqrt3\right)^2}$
=$\sqrt[6]{{1664-960\sqrt3}\over{832-480\sqrt3}}$
=$\sqrt[6]2$
