interesting problem

We simplify all of these expressions.

A:

$A=\frac{2^{\frac{1}{2}}}{4^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{(2^2)^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{2^{\frac{1}{4}}}\\ A=2^\frac{1}{4}$

B:

$B=\sqrt[12]{128}\\ B=\sqrt[12]{2^7}\\ B=2^\frac{7}{12}$

C:

$C=(\frac{1}{8^\frac{1}{5}})^2\\ C=(\frac{1}{(2^3)^\frac{1}{5}})^2\\ C=(\frac{1}{2^\frac{3}{5}})^2\\ C=(2^{-\frac{3}{5}})^2\\ C=(2^{-{\frac{6}{5}}})\\ $

D:

$D=\sqrt{\frac{4^{-1}}{2^{-1}\cdot8^{-1}}}\\ D=\sqrt{\frac{(2^2)^{-1}}{2^{-1}\cdot(2^3)^{-1}}}\\ D=\sqrt{\frac{2^{-2}}{2^{-4}}}\\ D=2$

E:

$E=\sqrt[3]{2^\frac{1}{2}\cdot4^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot(2^2)^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot2^-{\frac{1}{2}}}\\ E=\sqrt[3]{1}\\ E=1$

We then order them: C, E, A, B, D.