Is there any trick to quickly find the eigenvalues of this matrix? Is there any trick to quickly find the eigenvalues of this matrix:
$$
        \begin{pmatrix}
        33 & 6 & 6 \\
        6 & 24 & -12 \\
        6 & -12 & 24 \\
        \end{pmatrix}
$$
Besides being symmetric, the entries seem to be carefully chosen for the eigenvalues to be perfect squares.
By a trick I mean something like:

To find the the eigenvalues of
  $$
        \begin{pmatrix}
        11 & 6 \\
        15 & 2 \\
        \end{pmatrix}
$$
  note that the sum of entries in each row is 17 and get the second eigenvalue from the trace.

 A: You can symmetrically permute the matrix to centrosymmetric form by rotating the last row/column to the first:
$$
\begin{bmatrix}
24&6&-12\\
6&33&6\\
-12&6&4
\end{bmatrix}$$
Now a centrosymmetric matrix has even and odd eigenvectors with respect the middle index, so that means that your eigenvectors will be of the form $[\alpha,0,-\alpha]$ or $[\beta,\gamma,\beta]$ for scalars $\alpha$ and $\beta$. There is only one of the former, and two of the latter by symmetry considerations. So therefore it is easiest to try to find $\alpha$. Substituting it in as an eigenvector, we get $6\alpha = \lambda \alpha$, so $\lambda=36$ is the eigenvalue corresponding to the anti-symmetric eigenvector (which for the original matrix would be $(0,1,-1)$).
Substituting in the symmetric form, we get
$$ 3\begin{bmatrix}4 & 2 \\ 4 & 11\end{bmatrix}\begin{bmatrix}\beta\\\gamma\end{bmatrix}=\lambda \begin{bmatrix}\beta\\\gamma\end{bmatrix}$$
This $2\times 2$ problem is now very simple to solve.
A: Your matrix is:
$$\begin{pmatrix}33&6&6\\6&24&-12\\6&-12&24\end{pmatrix}$$
The eigenvalues are:
$$\lambda_1=33-(6+6)-(6+6)=9$$
$$\lambda_2=24-(-12)=36$$
$$\lambda_3=24-(-12)=36$$
So there's clearly something fishy going on here.
