Show that $\dfrac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =2^\frac{1}{6}$

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.

$$\frac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =\sqrt[6]{\frac{4^3(2-\sqrt3)^3}{4^2(3\sqrt3-5)^2}}=\sqrt[6]{\frac{4(2-\sqrt3)^3}{(3\sqrt3-5)^2}}=\sqrt[6]{\frac{4(8-12\sqrt3+18-3\sqrt3)}{(27-30\sqrt3+25)}}=$$ $$\sqrt[6]{\frac{4(26-15\sqrt3)}{(52-30\sqrt3)}}=\sqrt[6]{\frac{4(26-15\sqrt3)}{2(26-15\sqrt3)}}=2^{\frac{1}{6}}$$

Hint $ $ It arises simply by taking the $6$'th root of the lhs = rhs below

$$(8 - 4\sqrt 3)^{\large \color{#c00}3} =\, ((\sqrt 6 - \sqrt 2)^{\large \color{#0a0}2})^{\large \color{#c00}3} =\, ((\sqrt 6 - \sqrt 2)^{\large \color{#c00}3})^{\large \color{#0a0}2} =\, 2\, (12\sqrt 3 - 20)^{\large \color{#0a0}2}$$

As to your general question, yes, there are effective algorithms for deciding equality of real algebraic numbers. But they would not be easy to describe at the algebra-precalculus level. For one such algorithm see Renaud Rioboo, Towards faster real algebraic numbers.