Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$

The way to compute this is as follows: if $A$ is a matrix which has $\alpha$ as an eigenvalue and $B$ is a matrix that has $\beta$ as an eigenvalue, then $A \otimes I + I \otimes B$ is a matrix that has $\alpha + \beta$ as an eigenvalue. This is because if $Av = \alpha v$ and $Bw = \beta w$ then

$$ (A \otimes I + I \otimes B)(v \otimes w) = (Av \otimes Iw) + (Iv \otimes Bw) = (\alpha + \beta)(v \otimes w). $$

To get $A$ and $B$, you can use the companion matrix of a polynomial:

$$ \mathrm{Char. Poly.} \begin{pmatrix} 0 & 0 & \dots & -a_0 \\ 1 & 0 & \dots & -a_1 \\ 0 & 1 & \dots & -a_2 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & -a_{n-1} \end{pmatrix} = x^n + a_{n-1}x^{n - 1} + \dots + a_0. $$

If you do this computation, you'll find that the characteristic polynomial of $$ \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \otimes I_2 + I_3 \otimes \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix} $$ is $x^6 + 3x^5 - 4x^4 - 11x^3 + 25x^2 + 52x -39$ and that's the minimal polynomial of $\alpha + \beta$.

Here’s another method:

The roots of the second (quadratic) polynomial are $\beta$ and $\overline\beta=-1-\beta$ (because the sum of the roots has to be the negative of the next-highest coefficient).

Now take the polynomial $f(X)=X^3+X+1$, which is the minimal polynomial for $\alpha$, and write $g(X)=f(X-\beta)$, which is a polynomial with $\alpha+\beta$ for a root, coefficients in $\Bbb Q(\beta)$. Also, write $\overline g(X)=f(X-\overline\beta)$, just what you get from $g(X)$ by replacing $\beta$ by $\overline\beta$. If we’ve done everything right, we’ll get $g\overline g$ to have rational coefficients. In fact, \begin{align} g(X)&=X^3-3\beta X^2 +(10-3\beta)X +4-5\beta\\ \overline g(X)&=X^3+(3+3\beta)X^2+(13+3\beta)X+9+5\beta\\ g\overline g(X)&=X^6 +3X^5-4X^4-11X^3+25X^2+52X-39\,, \end{align} as desired.

(This is easily done by hand, if you work carefully. I confess that I was feeling lazy, and used a symbolic-computation package to do it. If working by hand, you may find it easier to use what you know about $\beta$, namely that it’s equal to $\frac{-1+\sqrt{13}}2$, when $\overline\beta$ is the same thing with a minus sign in front of the radical.)