A negative base is a point of conflict between the three commonly used meanings of exponentiation.
- For the continuous real exponentiation operator, you're not allowed to have a negative base.
- For the discrete real exponentiation operator, we allow fractional exponents with odd denominators, and
$$(-a)^{b/c} = \sqrt[c]{(-a)^b}= \left( \sqrt[c]{-a} \right)^b = (-1)^b a^{b/c} $$
(and this is allowed because every real number has a unique $c$-th root)
- For the complex exponentiation operator, exponentiation is multivalued. An exponentiation with denominator $n$ generally takes on $n$ distinct values, although one is generally chosen as the "principal" value.
For $(-5)^{2/3}$, these three exponentiation operators give
- Undefined
- $\sqrt[3]{25}$
- $\omega \sqrt[3]{25}$ is the principal value. The other two are $\sqrt[3]{25}$ and $\omega^2 \sqrt[3]{25}$, where $\omega = -\frac{1}{2} + \mathbf{i} \frac{\sqrt{3}}{2}$ is a cube root of $1$.
Unfortunately, which meaning of exponentiation is meant is rarely ever stated explicitly, and has to be guessed from context.
I'm guessing that the second one is meant.
In case you're curious, here is part of the rationale for the first and third conventions.
In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other negative. And it's not even clear that $(-5)^{\pi/5}$ should be meaningful!
For the third convention, the whole thing is like the idea of $\pm 2$ being the 'square root of 4', but for the fact the complexes cannot be cleanly separated into "negative" and "positive" to let us choose a specific one nicely.
A method is chosen for the principal value, based trying to get positive bases 'right' and trying to keep continuity as much as possible, but alas this convention gets the negative bases 'wrong'.
In some sense, this can be viewed as the principal value of $(-5)^{2/3}$ chosen to be "two-thirds of the way" from positive to negative.
The problem is that the law
$$b^{pq}=(b^p)^q$$
does not hold for $b<0$, so you are not allowed to use it.
Also note that an expression such as
$$(-27)^\frac23$$
can be interpreted in two different ways. See the thread How do you compute negative numbers to fractional powers? Neither of these two conventions saves the law $b^{pq}=(b^p)^q$.
One convention is to take
$$(-27)^\frac23 = \sqrt[3]{(-27)^2} = (-\sqrt[3]{27})^2$$
which gives $9$, and the other convention is to use principal value of the complex expression which (with usual choices of "principal") gives
$$9\left( -\frac12 + i\frac{\sqrt3}{2} \right)$$
or approximately $-4.5+7.79i$.
Let me emphasize again that neither convention of raising negative real numbers to rational exponents preserves the law $b^{pq}=(b^p)^q$.
Best Answer
${(a^b)}^c= a^ba^c$ is not an equation that holds for all $a,b,c$. Although that equation is true when all of $a,b,c$ are positive real numbers, as you've discovered it's not true in general.
The correct approach is generally considered to simplify the exponent and obtain the complex number, though there are actually three different ways you can think of exponents with a negative base, as discussed here.