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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.

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    $\begingroup$ The third equality is completely wrong... $\endgroup$ – Yanko Oct 21 '17 at 13:32
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    $\begingroup$ $(-27)^{1/3} = (-27)^{2/6} \neq ((-27)^2)^{1/6}$ $\endgroup$ – samjoe Oct 21 '17 at 13:36
  • $\begingroup$ @yanko I know it's wrong. I asked this question to understand why. $\endgroup$ – Daniele Oct 21 '17 at 13:48
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    $\begingroup$ It's obviously wrong because you reach a contradiction.. what made you think that would be true in the first place? $\endgroup$ – Yanko Oct 21 '17 at 14:58
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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?.

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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 $a>0$.

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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).

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