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.

We have
$$8 - 4\sqrt{3} = 2(4-2\sqrt{3}) = 2(3-2\sqrt{3}+1) = 2(\sqrt{3}-1)^2,$$
so $\sqrt{8-4\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$.
Then we see that looking at $(\sqrt{3}-1)^3$ is a good idea:
$$(\sqrt{3}-1)^3 = (4-2\sqrt{3})(\sqrt{3}-1) = 6\sqrt{3} - 10,$$
so $12\sqrt{3}-20 = 2(\sqrt{3}-1)^3$ and $\sqrt[3]{12\sqrt{3}-20} = \sqrt[3]{2}(\sqrt{3}-1)$.
$\frac{\sqrt{2}}{\sqrt[3]{2}} = 2^{1/6}$ is then easy to see.
