Why isn't $\sqrt[12]{\left(-1\right)^6}$ equal to $\sqrt{-1}$? Why isn't $\sqrt[12]{\left(-1\right)^6}$ equal to $\sqrt{-1}$? Clearly, the square root isn't defined. We have divided the index and the exponent by $6$. The theorem says that $$\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}$$ where $a\ge0$. How does it work when $a<0$, or it doesn't?
PP. I see that we can write $\sqrt[12]{(-1)^6}=\sqrt[12]{1^6}=1,$ but my question still holds.

$\sqrt[n]{a^m}$ says do calculations in this order:

*

*Find $a^m$.

*Find the greatest real number you can raise to the $n$th power to get that previous result.

$\sqrt[nk]{a^{mk}}$ says do calculations in this order:

*

*Find $mk$ and separately $nk$.

*Find $a$ raised to the first result from 1.

*Find the greatest real number you can raise to (1)'s second result,such that you get the result from 2.

My point is that the sequence of things that you do in the two expressions are very different according to purely the order of operations. So in some sense it should be surprising that they would ever end up giving you the same result.
And that's right: they don't give the same result. Except in certain conditions, it can be proved that they will give the same result. One of those conditions is when $a>0$, in which case it can be proved these two things lead to the same end. That proof relies on $a$ being positive. In another situation where $a<0$, maybe the two things will end up the same, maybe not.
