Minimal polynomial of $\alpha+\beta$ over $\Bbb Q$ where $\alpha$ is a root of $x^3 − 2$ and $\beta$ is a root of $x^2 + x + 1$. Let $\alpha$ be a root of the polynomial $x^3 − 2 \in \Bbb Z[x]$, and let $\beta$ be a root of the polynomial $x^2 + x + 1 \in \Bbb Z[x]$. Determine the minimal polynomial of $\alpha+\beta$ over $\Bbb Q$.
My thought is using tower law, we know that $[\Bbb Q(\alpha,\beta):\Bbb Q]=6$, and then proceed to showing $[\Bbb Q(\alpha + \beta):\Bbb Q(\alpha)]$ and $ [\Bbb Q(\alpha + \beta):\Bbb Q(\beta)]$ are both greater than $1$, thus showing $[\Bbb Q(\alpha + \beta):\Bbb Q]$ must be $6$ or more, therefore $\Bbb Q(\alpha + \beta)$ is the same as $\Bbb Q(\alpha,\beta)$, sharing the same minimal polynomial $(x^2 + x + 1)(x^3 − 2)$. 
Is that correct?
 A: Until the last sentence, it seems correct.
But, $\alpha+\beta$ is not a root for the polynomial $m_\alpha\,m_\beta$. Instead, find an equation that $x=\alpha+\beta$ satisfies, e.g.
$$(x-\beta)^3=2$$
Expand and use $\beta^2=-\beta-1$, then group the $\beta$'s together:
$$x^3-3x-3=(3x^2+3x)\beta \quad\quad (1)$$
unless I miscalculated. Now multiply $(1)$ by $(3x^2+3x)$ then square $(1)$ and add them to get rid of $\beta$, hence arriving to the equation (of degree $6$).
A: No. The minimal polynomial of $x$ over $F$ is the lowest-degree monic polynomial with coefficients in $F$ that has $x$ as a root. There is no such thing as a minimal polynomial of a field (whether over another or not). All you know by the tower law (since $gcd(2,3) = 1$) is that the minimal polynomial of $a+b$ is of degree $6$ because $a+b$ generates $\mathbb{Q}(a+b)$. To explicitly find it, just expand powers of $a+b$ and reduce modulo $a^3-2$ and $b^2+b+1$. Clearly you will get a non-trivial linear relation with sufficiently many powers, even if you didn't know the degree of the minimal polynomial.
