= Solution
{c}
Take the <Nilpotent Jordan block>
$$
\boxed{A=J_3(0)=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.}
$$
Then
$$
A^2=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\ne0,\qquad A^3=0.
$$
Suppose that a complex <matrix> $B$ of size three satisfied $B^3=A$. We would have $B^9=A^3=0$, so $B$ would be a <nilpotent matrix>. Every <eigenvalue> of $B$ must then be zero: if $Bv=\theta v$ with $v\ne0$, applying $B^9$ gives $\theta^9v=0$. Its <characteristic polynomial> is therefore $t^3$, and the <Cayley-Hamilton theorem> gives $B^3=0$. This contradicts $B^3=A\ne0$. Hence
$$
\boxed{A^2\ne0\quad\text{but }A\text{ has no complex matrix cube root}.}
$$
The obstruction is that a <nilpotent matrix> of size three cannot have <nilpotency index> greater than three.
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