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

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}
