Cube root of a matrix ($B^3=A$)

A cube root of a matrix $A$ is a square matrix $B$ of the same size over the specified field satisfying $B^3=A$. Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block $J_3(0)$ has none, even though its square is nonzero. If $B^3=J_3(0)$, then $B^9=0$, making $B$ a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives $B^3=0$, a contradiction.