TITLE: Find the minimal polynomial for $\sqrt[3]{2} + \sqrt[3]{4}$ over $\mathbb{Q}$
QUESTION [6 upvotes]: Find the minimal polynomial for $\sqrt[3]{2} + \sqrt[3]{4}$ over $\mathbb{Q}$
I havent covered galois theory, this is an exercise from the chapter algebraic extensions in gallian.
I can see that $\sqrt[3]{2} + \sqrt[3]{4}$ $\in \mathbb{Q}(2^{1/3})$, after this I am clueless.
Need hints will finish the proof.

REPLY [2 votes]: Let $\beta=\sqrt[3]{2}$ and $\alpha=\beta+\beta^2$.
A systematic way to find the minimal polynomial of $\alpha$ is to consider the map $\mu: x \mapsto \alpha x$ on $\mathbb Q(\beta)$, write its matrix with respect to the basis $1, \beta, \beta^2$, and then find the minimal polynomial of this matrix.
This works because $p(\mu)(x)=p(\alpha)x$ for every polynomial $p \in \mathbb Q[T]$.
The matrix is 
$\pmatrix{ 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 }$ and its characteristic polynomial is $x^3 - 6 x - 6$. Since this polynomial is irreducible over $\mathbb Q$ by the rational root theorem, it is the minimal polynomial.