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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.
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?.
The mistake here:
$$(-27)^{\frac{2}{6}}=(-27)^{\frac{1}{3}}.$$ It's wrong because $$(-27)^{\frac{2}{6}}=|-27|^{\frac{1}{3}}.$$ Also if we write $a^{\frac{1}{3}}$ then it's better to think that $a>0$.
you could aswell prove this:
$-1 = \sqrt {(-1)^2}=\sqrt{1}=1$
Clearly the first equality is wrong (and your's third equality is wrong for similar reason).
