<s>[INST] determinant of sum of anticommuting matrices Let $A, B \in \mathcal{M}_n(\mathbb{C})$ such that $AB=-BA$ and $det(A)=0$ and $det(A+B)=0.$
Prove that $det(A^3+B^3)=0$.
I don't really know how to approach this question.
 [/INST] You only need $AB=-BA$ and $\det(A+B)=0$, not $\det A=0$, to prove that $\det(A^3+B^3)=0$. The reason is that $A^3+B^3$ can be factored such that $A+B$ is a factor.
Using $BA=-AB$ and always normalising products to the form $A^pB^q$, i.e. all $A$ factors before all $B$ factors, the product
$$
(A+rB)(A+B)(A+sB)=A^3+(r+s-1)A^2B+(r+s-rs)AB^2+rsB^3 
$$
which equals to $A^3+B^3$ if $rs=1$ and $r+s=1$. This happens if $(r,s)=(z,\bar z)$ where $z=e^{\pm\pi i/3}=\frac{1\pm\sqrt{3}i}{2}$, so you get
$$
A^3+B^3=(A+zB)(A+B)(A+\bar zB)
$$
and therefore
$$
\begin{align}
\det(A^3+B^3)&=\det[(A+zB)(A+B)(A+\bar zB)]\\
&=\det(A+zB)\,\det(A+B)\,\det(A+\bar zB)=0.
\end{align}
$$
</s>