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

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