Determinant of sum of squares of commuting matrices

Question

I have the following question from a math competition, can anyone help me solve this:

Let $A,B\in M_n(\mathbb{R})$ be two commuting matrices ($AB=BA$). Prove that $\det(A^2+B^2)\ge0$.

Thanks in advance.

Answer 1

$det(A^2+B^2)=det((A+iB)(A-iB))=det(A+iB)det(A-iB)=det(A+iB)det(\overline{A+iB})=det(A+iB)\overline{det(A+iB)}\geq0$.