polynomials - Finding the eigenvalues of $A=left(begin{smallmatrix} a & 1 & 1 \ 1 & a & 1 \ 1 & 1 & a \ end{smallmatrix}right)$

I would like to calculate the eigenvalues of the following matrix $A$, but the factorization of the characteristic polynomial does not seem to be easy to compute.

$A=\pmatrix{
a & 1 & 1 \\

1 & a & 1 \\
1 & 1 & a \\
},\ a\neq1,\ a\neq-2$

$f(\lambda)$ = Char$(A,\lambda)$ = $(a-\lambda)^3-3(a-\lambda)+2 = -\lambda^3 + 3a\lambda^2 + 3\lambda(1-3a^2) + (a-1)^2(a+2)$

I have thought about using the Rational-Root Theorem (RRT), so possible roots of $f(\lambda)$ are $(a-1),\ (-a+1),\ (a+2),\ (-a-2)$, and much more, as for example in the case $a=2$ we should also test whether $f(\pm2)=0$ or not, am I wrong?

The eigenvalues of $A$ are $a-1$ and $a+2$ (computed with Wolfram Alpha). This result can be obtained using RRT, computing $f(a-1)$ and $f(a+2)$ and realizing that both are equal to zero but, is there an easier and 'elegant' way to find these eigenvalues?

Answer

Basically, you need to solve $(a-\lambda)^3-3(a-\lambda)+2 =0$ for $\lambda$. Don't expand the brackets, instead denote: $t=a-\lambda$. Then:
$$t^3-3t+2=0 \Rightarrow (t-1)^2(t+2)=0 \Rightarrow \\
t_1=1 \Rightarrow a-\lambda =1 \Rightarrow \lambda_1 =a-1\\
t_2=-2\Rightarrow a-\lambda =-2 \Rightarrow \lambda_2=a+2.$$