Compute the $n$-th power of the difference of the following matrices. Given the matrices $$A=\begin{pmatrix} a & c & b \\ c & b & a \\ b & a & c \end{pmatrix}$$ and $$B=\begin{pmatrix} b & a & c \\ a & c & b \\ c & b & a \end{pmatrix}$$
for $a,b,c \in \mathbb{R}$
find $$(A-B)^n$$ for $n \in \mathbb{N}$.
I can see that the matrices are symmetric and also they have the same determinant since $B$ can be obtained by swapping the rows of $A$ twice.
I attempted to solve this by using the Cayley-Hamilton theorem.
First I find the characteristic polynomial which is $$p(k)=-k^3+(a+b+c)k^2-(ab+bc+ac)k+3abc-b^3-a^3-c^3$$
which is the same for both matrices.
Now by Cayley-Hamilton theorem, I have $$-A^3+(a+b+c)A^2-(ab+bc+ac)A+3abc-b^3-a^3-c^3=0$$
and $$-B^3+(a+b+c)B^2-(ab+bc+ac)B+3abc-b^3-a^3-c^3=0$$
By subtracting those two equations by parts, I get $$-A^3+B^3+(a+b+c)(A^2-B^2)-(ab+bc+ac)(A-B)+3abc-b^3-a^3-c^3$$
and this is where I am stuck, I attempted to use the identities $$a^3+b^3+c^3-3abc=\frac{1}{2} (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
and $$2ab+2bc+2ca=(a+b+c)^2-a^2-b^2-c^2$$ but couldn't end up with something useful.
My question is, first of all, is my approach correct? How should I proceed?

Note that $M = A - B$ can be written in the form
$$
\pmatrix{
p&r&q\\
r&q&p\\
q&p&r},
$$
where $p+q+r = 0$. Verify that
$$
M \pmatrix{1\\1\\1} = 0,
$$
which means that $\det(A - B)$ must be zero. With that, we find the characteristic polynomial to be
$$
\det(M - kI) = -k^3 - (pq + pr + qr)k.
$$
Let $\alpha = pq + pr + qr$. The Cayley Hamilton theorem tells us that
$$
-M^3 - \alpha M = 0 \implies M^3 = -\alpha M.
$$
With that established, we can conclude that for $n \geq 2$ we have
$$
M^n = \begin{cases}
\alpha^{n/2 - 1} M^2 & n \text{ is even,}\\
\alpha^{(n-1)/2} M & n \text{ is odd.}
\end{cases}
$$
