Where is my mistake ?$ (-1)^3=-1 \to \sqrt[3]{-1}=-1 $

Question

Where is my mistake ? (In the field of real numbers)

$$ (-1)^3=-1 \to \sqrt[3]{-1}=-1 $$

$$\sqrt[3]{-1}=-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{(1)^2}=1$$

$$-1=^?1 $$

Answer 1

The mistake lies in assuming that $(1)^{\frac{1}{6}}=1$, this is because, $1$ has two sixth roots in the field of real numbers, $1,-1$

Answer 2

That is the same error as in $1=\sqrt{1}=\sqrt{(-1)^2}=-1$

That is based on the fact: $1^\frac{1}{2} = \sqrt{1} = \pm 1$
The same is true for $1^\frac{1}{6} = \sqrt[6]{1} = \pm 1$
So you get $-1=\pm 1$ and that is true.