algebra precalculus - What is wrong with this proof? -3 = 3

What is wrong with this proof?

$-3 = \sqrt[3]{-27} = {(-27)}^{\frac 13} = {(-27)}^{\frac 26} = \sqrt[6]{{(-27)}^2} = \sqrt[6]{{27}^2} = {(27)}^{\frac 26} = {(27)}^{\frac 13} = \sqrt[3]{27} = 3$

This is obviously false since $-3 \neq 3$.

But still I can't figure out which equation is the wrong one and why that is.

Thanks in advance for anyone who will help.

Answer

As suggested in comments, the $(-27)^{2/6} \neq ((-27)^2)^{1/6}$, because $a^{bc}=(a^b)^c$ does not generally hold for $a<0$.

You can find more related info in @mrf's answer here: Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?.