Find the minimal polynomial for $\sqrt[3]{2} + \sqrt[3]{4}$ over $\mathbb{Q}$ 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.
 A: 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.
A: Considering what you've said about $\mathbb Q(2^{1/3})$, cube should suffice. Calculate the powers up to 3rd, and do some linear combinations.
$$\begin{array}{c|c|c|c|}
& 1& \sqrt[3]2& \sqrt[3]4 \\ \hline
\alpha^0& 1& 0& 0 \\
\alpha^1& 0& 1& 1 \\
\alpha^2& ?& ?& ? \\
\alpha^3& ?& ?& ? \\
\hline
\end{array}$$

 $x^3-6x-6=0$

A: Let $x=a(1+a)$, where $a^3=2$.  Then
$$x^3=a^3(1+3a+3a^2+a^3)=2(1+3a(1+a)+2)=6x+6$$
Thus $x=\sqrt[3]2+\sqrt[3]4$ is a root of the (irreducible) cubic $x^3-6x-6=0$.
A: let $x=\sqrt[3]{2},\alpha=x+x^2$
$$
\alpha^3 = (x+x^2)^3 =x^3(1+x)^3=2(1+3x+3x^2+2)=6+6\alpha
$$
