Is $\sqrt[3]{-1}=-1$?

Question

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...

Answer 1

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.

Answer 2

The notation $\sqrt[3]{-1}$ is a little bit ambiguous, since there are exactly three third roots of $-1$ over the complex numbers (in general, there are exactly $n$ $n-$roots of any complex number $z$ so the notation $\sqrt[n]{z}$ is ambiguous too).

Since

$$(-1)^3 = -1$$

$-1$ is one of those roots, but there are other two, namely, the roots of the equation

$$x^2-x+1$$

Which arise from the factorization

$$x^3+1=(x+1)(x^2-x+1)$$