= Truncated binomial cube root of the identity plus a nilpotent matrix
{title2=$\left(\sum_{j=0}^{m-1}\binom{1/3}{j}N^j\right)^3=I+N,\quad N^m=0$}
Over the <complex numbers>, if $N^m=0$, then
$$
Q=\sum_{j=0}^{m-1}\binom{1/3}{j}N^j,\qquad Q^3=I+N.
$$
Here the generalized <binomial coefficient> is $\binom{\alpha}{j}=\alpha(\alpha-1)\cdots(\alpha-j+1)/j!$, with $\binom{\alpha}{0}=1$. This is an exact finite identity, even when an infinite <binomial series> would not converge.
One can construct the required <polynomial> $q(z)=1+\sum_{j=1}^{m-1}c_jz^j$ without any infinite series. At <polynomial degree> $j$, the <coefficient> of $q(z)^3$ is $3c_j$ plus terms involving only earlier <coefficients>. Recursively choose it to agree with $1+z$ through <polynomial degree> $m-1$; the first <coefficients> are $1/3,-1/9,5/81$. The stated generalized <binomial coefficients> solve this recursion: near $z=0$, the <binomial series> for $(1+z)^{1/3}$ cubes to $1+z$, and truncating it changes no <coefficient> below <polynomial degree> $m$. Thus $q(z)^3-(1+z)$ is divisible by $z^m$ as a <polynomial identity>. Substituting the <nilpotent matrix> $N$ kills the remaining terms, all divisible by $N^m$. For a <Jordan block> $\lambda I+N$ with $\lambda\ne0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar <cube root> of $\lambda$.
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