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I was surfing Facebook and I ran into this question posted by a high school student:

$$\text{Which value equals } (-1)^{\frac{1}3}?\quad\text{Is it } 1\text{ or }-1?$$

He said that he asked this because he did it in two ways, both of which seem valid but they generate different answers.

Way #1: $$(-1)^{\frac{1}3}=\sqrt[3]{-1}=-1$$ since $(-1)^3=-1$. However, he also did something else.

Way #2: $$(-1)^{\frac{1}3}=(-1)^{\frac{2}6}=\left((-1)^2\right)^{\frac{1}6}=\sqrt[6]1=1$$ and this solution seems valid as well. He's confused, and after reading his question, I became somewhat confused as well. I know that $1^3\ne-1$, but I can’t see why way#2 is wrong.

Which solution above is invalid? Or is $(-1)^{\frac{1}3}$ undefined? Thanks in advance. I apologize if this is a stupid question.

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    $(-1)^6=1$, so $\sqrt[6]{1}=-1$ ALSO. Your are mixing REAL roots with COMPLEX roots. You need to be careful. In a same way, it can be "prrof": $$i=\sqrt{-1}=(-1)^{1/2}=(-1)^{2/4}=\sqrt[4]{(-1)^2}=\sqrt[4]{1}=1$$ – Tito Eliatron Nov 14 '20 at 12:26
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    In my mind, I only define fractional exponents when the base is nonnegative, due to this problem. – littleO Nov 14 '20 at 12:28
  • $(-1)^{\frac{1}3}$ is not $1$ (cube both of them to see this). But it might be $-1$ or $\frac12 +i\frac{\sqrt 3}{2}$ or even perhaps $\frac12 -i\frac{\sqrt 3}{2}$ depending on your definition of rational roots of numbers which are not positive reals – Henry Nov 14 '20 at 12:30

3 Answers3

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The first solution is valid. Note that it is natural to define $(-1)^{1/3}$ as $\sqrt[3]{-1}$ and that indeed $-1$ is the only (real) cube root of $-1$.

However, the second solution is based upon the formula $a^{bc}=(a^b)^c$, which only holds for $a>0$.

José Carlos Santos
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Try cubing $1$ and you will see that it is not a cube root of $-1$. It is a sixth root of $1$ though. Others will comment more on the issues involved in defining roots.

Since cubing is bijective on the real numbers, a single real cube root can be defined for each real number.

What you have done is involve squaring too, and squaring real numbers is neither injective nor surjective (neither one-to-one nor onto). And that is why the second method does not work.

When you move to the complex numbers the question of Nth roots is different - each non-zero complex number has N different Nth roots, and you have to be really careful if you need to pick just one.

Mark Bennet
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The fact is that $$a^{mn}={(a^m)}^n$$ need not be always valid. The correct way is $$x={(-1)}^{1/3}$$ $$x^3+1=0$$ $$(x+1)(x^2-x+1)=0$$ if x is real $$x=-1$$

Albus Dumbledore
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