So if you try to solve $(-1)^{1/3}$ you can do
$(-1)^{1/3} = \sqrt[3]{-1} = -1$ (cubic root of $-1$)
But what if I write $1/3$ as $2/6$?
$$(-1)^{1/3} = (-1)^{2/6}$$
So $(-1)^{2/6} = \sqrt[6]{(-1)^2}$ (sixth root of $(-1)^2$)
So we have $\sqrt[6]{(-1)^2}= \sqrt[6]{1} = + 1$
Then we have: $+1 = -1$
How is that possible?
Exponentiation $a^b$ is not well-defined for all combinations of real numbers $a$ and $b$. Some of the cases where it has good definitions are
When $a$ is arbitrary and $b$ is a nonnegative integer.
When $a\ne 0$ and $b$ is any integer.
When $a\ge 0$ and $b$ is a positive rational number.
When $a\ge 0$ and $b$ is a positive real number.
When $a$ is $e$ and $b$ is an arbitrary complex number.
These gives rise to quite different definitions which fortunately happen to agreee on the result where the domains overlap.
Your case $(-1)^{1/3}$ is not in any of these domains, and there is no generally accepted well-behaved definition that extends the domain to $(-1,1/3)$. Your own calculation shows that such a definition cannot exist while also respecting the usual power rules.
... Well ... you could posit an ad-hoc definition that gives meaning to $a^b$ in the case that
But that still would not enable your calculation with $b=2/6$.
