TITLE: Is $\sqrt[3]{-1}=-1$?
QUESTION [5 upvotes]: I observe that if we claim that $\sqrt[3]{-1}=-1$, we reach a contradiction.
Let's, indeed, suppose that $\sqrt[3]{-1}=-1$. Then, since the properties of powers are preserved, we have: $$\sqrt[3]{-1}=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$$ which is a clear contradiction to what we assumed...

REPLY [8 votes]: You say

since the properties of powers are preserved

but this is not true, and you have given a proof that exactly the opposite is true:
There is no rule that, for $a$ negative, one has $a^{bc} = (a^b)^c$. Indeed, we can see that this is not a rule even without using cube roots:
$$
-1 = -1^{1/1} = -1^{2/2} ``=" (-1^2)^{1/2} = \sqrt{1} = 1.
$$
This rule is only true when $a$ is a positive real number.