Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore
\begin{align*}
(\gamma - \beta)^3 &= \alpha^3 = -\alpha - 1 \\
\gamma^3 - 3\gamma^2\beta+3\gamma\beta^2-\beta^3 &= -\alpha - 1
\end{align*}
and using $\beta^2+\beta-3=0$, i get $\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1$.
I think the minimal polynomial of $\alpha+\beta$ has degree $6$, then i was trying raise to the $2^{nd}$ power the last equation, but i don't get nothing to obtain the minimal polynomial of $\alpha + \beta$.

Note that

 
*
 
*The polynomial $a^3+a+1$ has three distinct roots, exactly one of which is real.$\\[4pt]$
 
*The polynomial $b^2+b-3$ has two distinct roots, both of which are real.
 
 
 By pairing one of the non-real roots of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct non-real roots.
 
 By pairing the real root of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct real roots.

Thus the minimal polynomial of $\gamma$ has degree at least $4$, so $[\mathbb{Q}(\gamma):\mathbb{Q}]\ge 4$.
 
 Continuing the algebra from where you left off . . .
 \begin{align*}
 &
 \gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1
 \\[4pt]
\implies &
\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -(\gamma-\beta) - 1
\\[4pt]
\implies &
\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}
\\[4pt]
\implies &
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)^2
+
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)
-
3
=
0
\\[4pt]
\implies &
\frac{\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39}{\left(3\gamma^2+3\gamma+5\right)^2}=0
\\[4pt]
\implies &
\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39=0
\\[4pt]
\end{align*}
so $\gamma$ is a root of the polynomial
$$p(x)=x^6+3x^5-4x^4-11x^3+25x^2+52x-39$$
hence $4\le [\mathbb{Q}(\gamma):\mathbb{Q}]\le6$.

But from the equation
$$\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}$$
it follows that $\beta\in\mathbb{Q}(\gamma)$, hence $[\mathbb{Q}(\gamma):\mathbb{Q}]$ must be a multiple of $3$.

Thus we must have $[\mathbb{Q}(\gamma):\mathbb{Q}]=6$, hence $p(x)$ is the minimal polynomial of $\gamma$.
