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

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.
