Findinf the remaining eigenvalues of a $3\times 3$ matrix Let $n$ be a fixed natural number. We wish to compute the eigenvalues of the matrix below:
$$\begin{bmatrix}
a&amp;b&amp;c\\
d&amp;e&amp;f\\
g&amp;h&amp;i
\end{bmatrix}$$
We have a condition on the entries: $a+b+c=d+e+f=g+h+i=n$. It follows that one of the eigenvalues is $n$. I want the other eigenvalues in terms of the entries.

For example, if $\lambda_1 , \lambda_2,n$ are the eigenvalues, then $\lambda_1 + \lambda_2=n - a-e-i$.
Like this, I want both $\lambda_1 , \lambda_2$ in terms of the entries, not the sum of them.
Since you also have $\det A = n \lambda_1 \lambda_2$ (where $\det A$ can be expanded to your entries) see that
$$\lambda_2 = n-a-e-i-\lambda_1$$
and thus
$$\det A = \frac{(n-a-e-i-\lambda_1)\lambda_1}n = \frac1n \lambda_1^2 + (1-\frac1n(a+e+i))\lambda_1$$
So you can solve the quadratic to get explicit formulas (the two roots will be $\lambda_1$ and $\lambda_2$. The quadratic is
$$\lambda^2 + (n-\mathrm{tr}(A))\lambda - n\det A = 0$$
so
$$\lambda_{1,2} = \frac{\mathrm{tr}(A)-n}2 \pm \sqrt{\frac{(n - \mathrm{tr}(A))^2}4 + n\det A}$$

Replacing $\det A$ and $\mathrm tr A$ with the formula in terms of $a,b,\ldots, i$ will technically give you an explicit formula only dependent on your constants, but it's going to be a pain to write down or use.
I think the $\lambda_1 \lambda_2 n$ should be equal to $(-1)^n (-1)^n \det(M)$. Now putting them in a quadratic equation may help a little bit.