Prove that $\det(A+B)\det(A-B)=0$

$$\begin{aligned}\det(A+B)\det(A-B)&=\det(A+B)\det(A^T-B^T) \\ &= \det(AA^T+BA^T-AB^T-BB^T)\\&=\det(BA^T-AB^T)\end{aligned}$$

Now use the assumption that $n$ is odd and the fact that $C=BA^T-AB^T$ is skew-symmetric: $C=-C^T$.