Given matrices $A$, $C$ such that $ACA=0$, show that characteristic polynomial of $AB$ and $A(B+C)$ is same for any matrix B.

Question

Given matrices $A$ and $C$ such that $ACA=0$, show that characteristic polynomial of $AB$ and $A(B+C)$ is same for any matrix B.This question was in a worksheet for Cayley Hamilton Theorem. The order of the matrices was not mentioned so I assume I have to prove it in general $A, B, C \in M_{n \times n}(\mathbb{R})$
If $r(A)=0$ then $A=0$ and the proof is trivial.
If $r(A)=n$ then $A$ is invertible so $C=0$ and the proof is trivial again.
So for the non trivial case, $1<r(A)<n$ which also implies $A$ is a matrix with $det(A)=0$
$r(AB)\leq r(A)$ and $r(A(B+C))\leq r(A)$ so both matrices must also have a zero determinant.

My attempt was to show charpoly $AB$ is satisfied by $AB+AC$ and vice-versa to conclude the charpoly must be equal (which now that I think about it, will not prove their characteristic polynomials are equal!!)
$(AB+AC)^n = (AB)^{n-1}A(B+C)$ for $n\in \mathbb{N}$
Now let $det(xI-AB)=f(x) = c_0 + c_1x + ... + c_{n-1}x^{n-1} + x^n$
Evaluate $f(AB+AC) = c_0I + c_1(AB+AC) + ... + c_{n-1}(AB+AC)^{n-1} + (AB+AC)^n$
$f(AB+AC) = c_0I + c_1(AB+AC) + ... + c_{n-1}(AB)^{n-2}A(B+C) + (AB)^{n-1}A(B+C)$
$f(AB+AC) = f(AB) + c_1AC + c_2(AB)AC + ... + c_{n-1}(AB)^{n-2}AC + (AB)^{n-1}AC$

And this is where I was stumped. Any hints on how to approach this question? Or just the solution straight up also works. Thank you.