Why is $((-8)^2)^{1/6} > 0 \text{ and } -2 = (-8)^{1/3}$?

Question

Why is $((-8)^2)^{1/6} > 0 \text{ and } -2 = (-8)^{1/3}$?

Doesn't this contradict the exponentiation rule (power of power)?

Answer 1

No, it doesn't.

Note that $\left(a^b \right)^c = a^{bc}$ requires that $a > 0$. From Wikipedia:

the identity

$\left(b^{r}\right)^s = b^{r \cdot s}$

cannot be extended consistently to cases where $b$ is a negative real number

Someone who knows about complex numbers more than I do can explain what happens in the case where $a < 0$.

Answer 2

The exponent rule you are thinking of is $(x^a)^b=x^{ab}$. This is only really valid for $x>0$. With some negative $x$, there are some values of $a,b$ where it happens to still be true. But in general, exponent rules do not apply when there is a negative base.

Also, $(-8)^{1/3}$ is not universally evaluated to $-2$ as you say. It depends on definitions of exponentiation. Sometimes such things (with a negative base) are left undefined because of this issue. Some computing software would evaluate $(-8)^{1/3}$ to be a certain non-real complex number in the first quadrant. Try entering (-8)^(1/3) into WolframAlpha for example, and see how it is not the same as cuberoot(-8).

Answer 3

Consider for instance $x^{2/3}$. Since $2/3=4/6$, it sounds reasonable to say that $x^{2/3}=x^{4/6}$. However, if we let $x$ be any real number (positive or negative) and use the "power of power" rule freely, we get:

$x^{2/3}=(x^2)^\frac{1}{3}$ is defined for every $x$

$x^{2/3}=(x^{\frac{1}{3}})^2$ is defined for every $x$

$x^{4/6}=(x^4)^\frac{1}{6}$ is defined for every $x$

$x^{4/6}=(x^\frac{1}{6})^{4}$ is not defined for every $x$

Hence we should conclude $x^{2/3}\neq x^{4/6}$ since both functions are disagreeing. So we have to chose between saying the rule doesn't always apply and $x^{2/3}=x^{4/6}$ for all $x$ or we can say that it always apply and there are some $x$ for which $x^{2/3}\neq x^{4/6}$. Since the latter is not practical, we chose to restrict the use of the exponent rule.

Answer 4

The problem is that there is a branch cut discontinuity in the complex plane. If you consider the square root function, and the following series of "equalities": $$1=1^2=\sqrt{1}\cdot\sqrt{1}=\sqrt{1\cdot 1}=\sqrt{(-1)(-1)}=\sqrt{-1}\cdot\sqrt{-1}=i^2=-1, $$ where is the error? It occurs here: $\sqrt{(-1)(-1)}\not=\sqrt{-1}\cdot\sqrt{-1}$. It comes down to this: you're approaching the branch cut discontinuity from two different directions, and you can't assume that the limiting values are equal.