How does $(x^3 + 8)^{-2/3} = 0$ become $(x^3 + 8) = 0$? Can a negative exponent always be simply dropped? It was explained to me that if:
$$(x^3 + 8)^{-2/3} = 0$$
then
$$(x^3 + 8) = 0$$
...  In other words, the negative fractional exponent can simply be dropped from the expression. What are the exact rules that apply to this statement?
Can a negative exponent ALWAYS be simply dropped from the expression or are there more stipulations that need to be accounted for when making the determination?
 A: Someone explained something to you incorrectly. And worse, they led you to believe something that is false is true.
If $$Q^{-2/3}=0$$ then recognize what negative exponents indicate: $$\frac{1}{Q^{2/3}}=0$$
So $1$ was divided by something and the result was $0$. There is no number $Q$ where this happens.
So your equation has no solution, unlike the other equation $(x^3+8)=0$ which does have a solution.
A: Well, the negative sign can validly be dropped from the exponent if the exponent identically equals $0,$ or if the base or the base raised to the exponent identically equals $1,$ or if the base identically equals $-1$ and the exponent is an integer.
But, as suggested by Ivan in the comments, such an inelegant rule is less useful than simply understanding the exponent laws properly.
P.S. $(x^3 + 8)^{-2/3} = 0$ does not imply $(x^3 + 8) = 0$, since the the latter equation (having $3$ solutions) fails to capture the fact that the former has no solution.
P.P.S. Conversely, $(x^3 + 8) = 0$ does not imply $(x^3 + 8)^{-2/3} = 0$ either, since the former's solution set is not a subset of the latter's.
