What is the correct answer for simplifying $\sqrt{-2}\,\sqrt{-2}\,\sqrt{-2}$?

Question

I saw this question in an online forum. Kinda curious since I have seen two good answers that technically both make sense.

$$\sqrt{-2}\,\sqrt{-2}\,\sqrt{-2}$$

Does anyone know what the correct, fully simplified answer is?

Answer 1

On the principal branch of the square root function, $\sqrt{-2}=i\sqrt{2}$. Therefore $(\sqrt{-2})^3 = i^3(\sqrt{2})^3=-i(\sqrt{2})^3$.

Interestingly, you cannot multiply the $-2$ together under the root because the "usual rule" $\sqrt{a}\sqrt{b}=\sqrt{ab}$ works only for positive real numbers. To see why this could cause problems, consider the following "proof" that $-1=1$ (taken from Wikipedia): $$-1 = i\cdot i = \sqrt{-1}\cdot\sqrt{-1} = \sqrt{(-1)\cdot(-1)} = \sqrt{1} = 1.$$

Answer 2

This is $\sqrt2i$ multiplied to itself three times. Thus we get $-2\sqrt2i$. Note that we aren’t allowed to simplify this into one radical since the radicands are negative.

Answer 3

Here the main issue is why many say, you can't apply rules of indices for a product of negative numbers. If so can you apply rules of indices for a product of imaginary numbers and then how can you explain the proof of De Moivre's theorem in the case of rational powers?
If you find square root of $ i$ or square root of $5+12i$ using De Moivre's theorem you get two roots in each case and we treat those as square roots of that particular number. Now if you apply the same approach to find square root of $1$,you get $ 1 , -1$ as square roots but we are used to consider only $1$ as the principal root.
If you take square root of $-2$ as $ 2^1/2i $ and multiply by it self you get the result $-2$ but if you multiply $-2 $ by it self and finally get the square root answer will be $ 2$ . If so how can you consider a particular approach is correct when you multiply three times by itself.
Therefore I think better way is to accept there can be more than one root when simplifying fractional powers of complex numbers in any form.