Raising a real number to a rational power is very simple, right? Consider the following example:
$$−27 = (−27)^{\frac{2}{3}\frac{3}{2}} = ((−27)^{\frac{2}{3}})^\frac{3}{2} = 9^\frac{3}{2} = 27$$
The issue arose because of this part: $-27^{\frac{2}{3}}$. Instead of taking the negative cube root, we first raised it to second power to get $729$ and then calculated $\sqrt[3]729=9$.
Fine, but shouldn't it be like guaranteed by the definition of raising a real number to rational power, that no matter what order of operation I choose, I get the same (correct) answer? What if it was used in some proof? That would immediately invalidate it, but how certain can we be that proofreaders remembered about this issue?
Wikipedia describes it not as something that should never be done, instead it is claimed one needs extra care when raising negative numbers to rational powers. If you think it should never be done, then I guess the article needs to be corrected.
Best Answer
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 $b^{pq}=(b^p)^q$.