What does $\sqrt[6]{-1}$ mean on wolfram? To solve $X^4-X^2+1=0$, wolfram wrote $\pm\sqrt[6]{-1}$ and $\pm(-1)^{5/6}$. Does it mean $\sqrt[6]{i}$ and $i^{5/6}$ or $\pm e^{\frac{i\pi}{6}}$ and $\pm e^{\frac{5i\pi}{6}}$ ? 
In fact, nothing looks to be good since solution are $e^{\pm i\pi/6}$ and $e^{\pm 5i\pi/6}$, so why such notation ?
Here the link
 A: Wolfram means
$$\pm\sqrt[6]{-1}=\pm\sqrt[6]{e^{i\pi}}=\pm e^{\frac{\pi i}6}$$
or in other words
$$\pm\sqrt[3]{\sqrt[2]{-1}}=\pm\sqrt[3]{i}=\pm\sqrt[3] {e^{\frac{i\pi}{2}}}=\pm e^{\frac{\pi i}6},$$
as Wolframalpha takes the principal branch $(-\pi,\pi]$, and so $i=e^{\frac{i\pi}{2}}$ and $-1=e^{i\pi}$ when computing roots.
For the divergence of solutions, just note that 
$$-e^{\frac{i\pi}{6}}=e^{-\pi i}\cdot e^{\frac{i\pi}{6}}=e^{-\frac{5i\pi}{6}}$$
and so on. So the solutions are indeed the same.
A: I assume the reason why Mathematica prefers $(-1)^{1/6}$ to $e^{i\pi/6}$ is because it has fewer transcendental constants, and is thus easier to work with (as a CAS). It's also simpler as a Mathematica expression, that is Power[-1, 1/6] as compared to Exp[Times[0 + 1/6 I, Pi]]; since the rational numbers are treated as atomic (as well as "complex rationals", i.e. elements of $\Bbb Q(i)$), that's one operation instead of two, which I'm sure is a factor in Simplify handling.
Mathematica uses the convention $a^b=e^{b\log a}$, where $\log$ has a branch cut at the negative reals. It follows from this definition that $(-1)^{1/2}=i$, $(-1)^{1/3}=e^{i\pi/3}$, and $(-1)^{1/6}=e^{i\pi/6}$ (and in general $(-1)^x=e^{i\pi x}$ for $x\in(-1/2,1/2]$).
The negative of $(-1)^{1/6}$ is $$-(-1)^{1/6}=e^{7i\pi/6}=e^{-5i\pi/6}=(-1)^{-5/6},$$ and the conjugate is $$\overline{(-1)^{1/6}}=e^{-i\pi/6}=e^{5i\pi/6}=(-1)^{5/6}.$$
A: Wolfram is right,
$$x^4-x^2+1=\frac{x^6+1}{x^2+1}$$ and the roots are those of $x^6+1$, but those of $x^2+1$.
But you are wrong to write $\sqrt[6]{-1}=\sqrt[6]{i}$ (as $-1\ne i$). Instead, $\sqrt[6]{-1}=\sqrt[3]{i}$, at least for some root of $-1$ and $i$.
