= Solution
{c}
We construct a <cube root of a matrix> on each <Jordan block>. Since $A$ is an <invertible matrix> over the <complex numbers>, its <Jordan normal form> is
$$
A=PJP^{-1},\qquad J=\bigoplus_{r=1}^{\ell}J_{m_r}(\lambda_r),\qquad \lambda_r\ne0.
$$
Consider one block $J_m(\lambda)=\lambda I+N$, where $N=J_m(0)$ is a <Nilpotent Jordan block> and $N^m=0$. Choose a complex <cube root> $\beta$ of $\lambda$, so $\beta^3=\lambda$.
We need a <polynomial> $q(z)$ with <constant term> one satisfying
$$
q(z)^3\equiv1+z\pmod{z^m}.
$$
This can be constructed using only <polynomial multiplication>. Write $q(z)=1+\sum_{j=1}^{m-1}c_jz^j$. Once $c_1,\ldots,c_{j-1}$ have been chosen, the <coefficient> of $z^j$ in $q(z)^3$ is $3c_j$ plus an expression depending only on those earlier <coefficients>. Choose $c_j$ to make this <coefficient> one at $j=1$ and zero at $2\leq j<m$. Division by three is possible over the <complex numbers>, so the recursion always succeeds. Its first <coefficients> are
$$
c_1=\frac13,\qquad c_2=-\frac19,\qquad c_3=\frac5{81}.
$$
This is the <truncated binomial cube root of the identity plus a nilpotent matrix>; equivalently $c_j=\binom{1/3}{j}$, where the generalized <binomial coefficient> is $\alpha(\alpha-1)\cdots(\alpha-j+1)/j!$ and the empty product is one.
Substitute $z=N/\lambda$. The <polynomial remainder> is a multiple of $N^m$, hence zero. It follows that
$$
C_m(\lambda)=\beta q(N/\lambda)
\quad\Longrightarrow\quad
C_m(\lambda)^3=\lambda(I+N/\lambda)=J_m(\lambda).
$$
For $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the <block diagonal matrix> $C=\bigoplus_r C_{m_r}(\lambda_r)$. Taking <matrix powers> respects these blocks, and conjugating back gives
$$
\boxed{B=PCP^{-1},\qquad B^3=PC^3P^{-1}=PJP^{-1}=A.}
$$
Thus every <invertible matrix> over the <complex numbers> has at least one <cube root of a matrix>. The construction uses only finitely many <matrix powers>.
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