Can anyone explain the mathematical, geometric or physical meaning of $(-1)^{1/3}$ Suddenly, I was very confused on the mathematical or geometric meaning about cube root of a minus real number, say
$(-1)^{1/3}$. In my opinion, I think $(-1)^{1/3}=-1$, because $(-1)^3=(-1).(-1).(-1)=-1$. But when I ask Mathematica with
N[(-1)^(1/3)]

I got
(*0.5 + 0.866025 I*) 

If the number is somewhat more complicated, say $\left(-\frac{27}{40}\right)^{1/3}$, how do I consider its sign, could it be considered as a positive number?
Or there is no plus or minus characteristic for this kind of number? Can I regard the cube root of a minus number as a minus one?
Sorry, if my question is too silly, kindly teach me. Thanks!
 A: Typically, we do algebra in the real numbers ( $\mathbb{R}$), but here you're looking at a different set. You are getting different answers because $\mathbb{C}$ is algebraically closed. Namely, the degree of a polynomial corresponds to its number of roots. Interestingly enough, this is not the case in the real numbers ( take $x^{2}+1=0$ for example, which is solving for the roots of a parabola that does not intersect the x-axis.) In other words:
$$(-1)^3=-1.$$
However, $e^{i \pi/3}$ and $e^{-i\pi/3}$ are also legitimate solutions in the complex numbers.
This is true because $e^{i \pi/3}=\cos(\pi/3)+i\cdot \sin(\pi/3)\approx .5+i\cdot.86603$
This is a solution because $(e^{i \pi/3})^{3}=e^{i \pi}=-1$.
Hence the famous equation: $e^{i \cdot \pi}+1=0$.
This is part of what makes mathematics in the complex numbers so well-behaved.
A: In complex numbers every number, not equal to zero, has n, n-th roots. All distinct.  The three cube roots of -1 are: $-1, 1/2 + \sqrt{3}/2 i, 1/2 - \sqrt{3}/2 i$. 
So there is room for ambiguity in which of these should be $(-1)^{1/3}$.   In general for $b >0$ $b^{1/3}$ is defined to be the specific and unique positive root which always exist and of which there is always only one.  But if $b$ is not a positive real (or zero) then there isn't  any positive real root. There is some, but not universal, consensus as to which of the n-th roots will get the honor of being defined $b^{1/n}$.  (Others will correct me if I am wrong.)
Now one thing about n-th complex roots of $b$.  There are always n of them and they  always are positioned on the complex plane so that they are inscribed in a circle around the origin, 0, with radius $\sqrt[n]{|w|}$ and each equal angles apart.
So if there is a convention.  It is usually to be the first one, one encounters on the circle when looking in a counter clockwise direction starting from the real axis.  That way if $b$ is positive real the first we'd encounter is the real root.
If we follow this convention then the first 3rd root of -1 that we encounter is $1/2 + \sqrt{3}/2 i$.  So that is why mathematica gave you what it did. (0.5 + 0.866025 I) is one cube root and it is the first one, one comes to in a counter-clockwise direction.
====  more advanced  (slightly) =====
If we look at a complex number $a + b i$ in the complex plane, we can view it as an ordered pair.  It is the point that is $a$ over on the x (real)  axis and $b$ up on the y (imaginary) axis.
But we can also not that if we view it as a ray or vector form the origin, 0, we can see it has a length $r = \sqrt{a^2 + b^2}$ and it has an angle $\theta$ measured from the real axis.  
So we can alternatively view the complex number as $r(\cos \theta + i \sin \theta)$.  It's the same number as $a = r\cos \theta$ and $b = r\sin \theta$.  But we are viewing it, instead of a rectangular coordinated point, we are iewing it as polar coordinated point.
There is a surprising benefit to viewing it this way:
If we let $w = e^{\ln r}(cos \theta + i sin \theta)$ is one complex number and $z = e^{\ln t}(cos \phi + i sin \phi)$ is another then $wz = e^{\ln r + \ln t}([\cos \theta \cos \phi - sin \theta \sin phi] + i [ cos \theta \sin \phi + \cos \phi \sin \theta]) = e^{\ln r + \ln t}(\cos (\theta + \phi) + i \sin (\theta + \phi))$.  
In other words we found a way to do multiplication but adding the angles and multiplying the lengths.
(We go a step further and define and write  $e^{a + b i} = e^a(\cos b + i \sin b)$ but we won't today.)
So anyway.  This means:  To find the n-th roots of $a + bi$ you convert it to $r(\cos \theta + i \sin \theta)$.  Then... wait for it... the n-th roots will be: ${r^{1/n}(\cos (\theta/n + 2k\pi/n) + i \sin (\theta/n + 2k\pi/n)} $ for $k = 0, 1, ..., n-1$. (as raising those to the $n$th power will result in $({r^{1/n}}^n(\cos (n\theta/n + 2kn\pi/n)  + i \sin (n\theta/n + 2kn\pi/n)=  r(\cos(\theta + 2k\pi) + i \sin (\theta + 2k\pi)) = r(\cos \theta + i \sin \theta)$.)
Which is why the convention is the define the default i-th root by the formula above with k = 0.
So if $-1 = 1(cos \pi + i \sin \pi)$ then $-1^{1/3} = 1^{1/3}(\cos \pi/3 + i \sin \pi/3) = 1/2 + \sqrt{3}/2 i$.
But the other two are $(\cos (\pi/3 + 2\pi/3) + i \sin (\pi/3 + 2\pi/3)) = \cos \pi + i \sin \pi = -1$
And $\cos (\pi/3 + 4\pi/3) + i \sin (\pi/3 + 4\pi/3) = \cos -\pi/3 + i \sin -\pi/3 = 1/2 - \sqrt{3}/2 i$.
A: You are right, $(-1)^3=-1$, hence $-1=(-1)^{1/3}$.
Anyway, there are two other solutions in complex numbers, $\dfrac{1\pm i\sqrt3}2$. Indeed,
$$\left(\frac{1\pm i\sqrt3}2\right)^3=\frac{1\pm i3\sqrt3-3\left(\sqrt3\right)^2\mp i\left(\sqrt3\right)^3}8=-1.$$
Mathematica prefers one of these.

Interpretation:
In the reals, multiplying by $-1$ is a mirroring around the origin. Three mirrorings amount to a single one.
In the complex, numbers are represented by points in a plane, and multiplying by a complex number is the combination of a rotation and a scaling, around the origin.
The two complex solutions correspond to rotations of $\pm60°$, with unit scaling. Hence three rotations are equivalent to a single one by $\pm180°$, i.e. multiplication by $-1$.
