Prove that $\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0 $

Question

Suppose that A, B and C are 2x2 matrices that switch between each other. Prove that $$\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0. $$

I did

$$A^3+B^3+C^3-3ABC=\frac12(A+B+C)((A-B)^2+(A-C)^2+(B-C)^2)$$

So, this determinant is equivalent to

$$\frac14[\det(A+B+C)]^2\det((A-B)^2+(A-C)^2+(B-C)^2)$$

But how can I prove that

$$\det((A-B)^2+(A-C)^2+(B-C)^2)\geq0?$$

Can someone help me? Thanks for attention.

Answer 1

Based on Aryaman Maithani's idea:

Presumably the matrices are real. Let $X=A-B,\,Y=B-C$ and $Z=C-A$. Then $X,Y,Z$ commute and $X+Y+Z=0$. Let $\omega$ be a primitive cube root of unity. Then \begin{aligned} \det(X^2+Y^2+Z^2) &=\det(X^2+Y^2+(-X-Y)^2)\\ &=4\det(X^2+Y^2+XY)\\ &=4\det\left[(X-\omega Y)(X-\bar{\omega}Y)\right]\\ &=4|\det(X-\omega Y)|^2\\ &\ge0. \end{aligned}