Question: If $A$ and $B$ are $n \times n$ matrices such that $A^{2} B^{2} = 0$ then which of the following is true?

a) $B^{2} A^{2} = 0$

b) $AB = 0$

c) Either $A^{2} =0$ or $B^{2} = 0$.

d) Either $A$ or $B$ has zero determinant.

My current reasoning:
$$A^{2} B^{2} = 0. $$
So, 
$$\det(A^{2}B^{2}) = 0$$
$$\det(A^{2}) \det(B^{2}) = 0.$$
 So from here either $A^{2}$ or $B^{2}$ is singular.

Also 
$$\det(A^{2} B^{2}) = \det(A)\det(AB)\det(B)$$

So, either $A$ or $AB$ or $B$ is singular.

I'm confused. I could only do this so far. 
Option c is incorrect as finding counter example for c was easy.
Is there any other way to find solution without using counter examples for incorrect options.

Answer: 
The inequality of arithmetic and geometric means implies that the product can never be larger than $6^3 = 216$. As a rule of thumb, the further you go away from the equal case, the smaller the product gets. So you only need to check the extreme cases, i.e. $(1,2,15)$ here.
Assume (x+y+z)/3=6 thus x+y+z =18. The largest of x,y, and x can be at most 15. (15+2+1)=18. This gives us a volume, (xyz), of 30 units cubed. This is also the smallest volume you can get. This can be visualized in 3D space. Every other combination would have three non zero terms. I haven’t worked out a proof of this but I assume it’s trivial although possibly tedious. I don’t know how much help this will be. But you don’t need to test all the other possibilities.