Find the $n$th power of a 3-by-3 circulant matrix Consider the matrix given as $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}$$
Write down a formual for $A^n$ for $n\in\mathbb{N}$.
$$$$
My attempt: The first that comes to mind is to diagonalize it and hence find the formual for $A^n$, but that is very messy, so I tried to do something else it goes as:
bserve that $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}=a_0\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+a_1\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}+a_2\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix}$$
Let $U=\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}$ then we will have that
$$\begin{matrix}U^2=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix} & \text{ and } & U^3=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\end{matrix}$$
Hence we have $A=a_0I+a_1U+a_2U^2 = a_0U^3+a_1U+a_2U^2=(a_0U^2+a_1I+a_2U)U$
This got me thinking that there might be an easy way to solve the above problem but, I was not able to make any further progress.
Please Help and thanks in advance.

I think you've reached a decent result and by just using the multinomial theorem you can obtain a generalized solution.
$A=a_0I+ a_1U + a_2 U^2 \\
A^n = (a_0I+ a_1U + a_2 U^2)^n\\$
Then the coeffecients of $U^3$, $U^2$ and $U^1$ are determined using the multinomial theorem and just plugged back into the result matrix.
$$
\begin{matrix}
\sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\
\end{matrix}
$$
The answer may seem obvious but unfortunately the multinomial expansion can't be simplified further (as per my knowledge). Even if it could it would just be a huge mess.
