What's the problem with $-2=(-8)^{\frac{1}{3}}=(-8)^{\frac{2}{6}}=\sqrt[6]{(-8)^{2}}=2$?

Question

$-2 =(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^{2}}=2$
first glance I want to show for $a^{rs}$ to work, both $a^r$ and $a^s$ need to be valid, but as you can see, both $-8^{\frac{1}{3}}, -8^{\frac{2}{2}}$ is defined. so I wanna know how do I prove the above statement is wrong

Answer 1

The sixth root of a number is considered even and for any even root, the answer can be positive or negative. You pick which one is best fit or both, which in this case is -2 as shown from your starting point.

Another issue is by stating that -2=2, you imply that 0=4 or 0=1, which is a tell-tale sign something is wrong.