A²+B² is singular when A³ = B³ and A²B = B²A

Idea. We find a nonzero matrix $A-B$ that kills $A^2+B^2$ from the right; then any nonzero column of $A-B$ lies in the kernel.

Step 1: expansion.
$$(A^2+B^2)(A-B)=A^3-A^2B+B^2A-B^3.$$
Group the terms: $(A^3-B^3)+(B^2A-A^2B)$. The first bracket is $0$ because $A^3=B^3$, the second is $0$ because $A^2B=B^2A$. Hence
$$(A^2+B^2)(A-B)=0.$$

Step 2: a nonzero vector in the kernel. Since $A\ne B$, the matrix $A-B$ is not zero, so it has a nonzero column; call it $v=(A-B)e_j\ne0$. Then
$$(A^2+B^2)v=(A^2+B^2)(A-B)e_j=0.$$

Step 3: conclusion. $A^2+B^2$ has a nonzero vector in its kernel, so it is not invertible, i.e. $\det(A^2+B^2)=0$.

Remark. Combined with the previous problem this is the classical Putnam statement: in the Putnam version one concludes $\det(A^2+B^2)=0$ using exactly this factorisation trick. Note that no commutation $AB=BA$ is assumed; only the two cubic relations.

Lesson. To prove a matrix is singular, look for a nonzero matrix that it kills on one side.