Find values that k may take if $k=\det\left(A^3+B^3+C^3\right).\det\left(A+B+C\right)$

Question

Let $A,B,C$ be $n\times n$ matrices with real entries such that their product is pairwise commutative. Also $ABC=O_{n}$. If $$k=\det\left(A^3+B^3+C^3\right).\det\left(A+B+C\right)$$ then find the value or the range of values that $k$ may take.

My Attempt

I tried $k=\left(\det(A+B+C)\right)^2\left(\det(A^2+B^2+C^2-AB-BC-CA)\right)$. but couldn't go further than this

Answer 1

$k=det(A^3+B^3+C^3).det(A+B+C)$$=det(A^3+B^3+C^3-3ABC).det(A+B+C)$

$=det\{(A+B+C)(A+\omega B+\omega^2 C)(A+\omega^2 B+\omega C)\}.det(A+B+C)$

$=(det(A+B+C))^2det(A+\omega B+\omega^2 C)det(\overline{A+\omega B+\omega^2 C})$

$=(det(A+B+C))^2(X+iY)(X-iY)$

$=(det(A+B+C))^2(X^2+Y^2)$

$\geq 0$

Answer 2

Steps:

1. Show that $\det (A^2 + B^2 + C^2 - AB - BC - CA ) \geq 0$. (See the linked problem.)
2. Show that $ k \geq 0 $.
3. For any $ k \geq 0$, find matrices $A, B, C$ that satisfy the condition.

Hints for the construction:

Diagonal matrices work.

The non-zero values are all the same, and equal to ...

The condition $ABC = 0$ means ...

If $ k > 0$, what does that imply about $A+B+C$?