Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find 
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0 
\end{align}
I tried to do similar tings such as 
\begin{align}
(x-a)^3 = (\sqrt[3]{2}(b+c\sqrt[3]{2}))^3 = 2 (b+c\sqrt[3]{2})^3
\end{align}
but this does not seems good. 
From the argument of 
\begin{align}
x^3-2 = (x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)
\end{align}
where $w^2+w+1=0$, 
I came up with some idea, and do the computation via mathematica. And i figure out the minimal polynomial for that is 
\begin{align}
&(x-(a+b\sqrt[3]{2}+c\sqrt[3]{4}))(x-(a+bw\sqrt[3]{2}+cw^2\sqrt[3]{4}))(x-(a+bw^2\sqrt[3]{2}+cw\sqrt[3]{4}))  \\
&=x^3 +3 a^2 x-3 a x^2-6 b c x-2 b^3-4 c^3 +6 a b c-a^3
\end{align}
The question is how to obtain this polynomial from the starting point. 
 A: An alternative approach.
Unless $b=c=0$, $\alpha=a+b\sqrt[3]2+c\sqrt[3]4$ has degree $3$ and so
a cubic minimum polynomial. Let's assume this. Observe that
$$\pmatrix{a&b&c\\2c&a&b\\2b&2c&a}\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}=\alpha\pmatrix{1\\\sqrt[3]2\\\sqrt[3]4}.$$
Therefore $\alpha$ is an eigenvalue of
$\pmatrix{a&b&c\\2c&a&b\\2b&2c&a}$. Its characteristic polynomial is
therefore the minimum polynomial of $\alpha$.
A: Since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$, we obtain:
$$a-x+b\sqrt[3]{2}+c\sqrt[3]{4}=0$$ or
$$(a-x)^3+2b^3+4c^3-6(a-x)bc=0,$$
which is third degree.
I hope now it's clear.
By the way, we got your polynomial exactly. 
A: I don't get it, where is a problem, a solution is direct. Lift the starting equation
$$x-a=b\sqrt[3]{2}+c\sqrt[3]{4}$$
on $3$, so have:
$$(x-a)^3=(b\sqrt[3]{2}+c\sqrt[3]{4})^3$$
so 
$$ (x-a)^3=2b^3+4c^3+6bc(\underbrace{b\sqrt[3]{2}+c\sqrt[3]{4}}_{x-a})$$
and we are done:
$$ p(x)=(x-a)^3-6bc(x-a)-2b^3-4c^3$$
So we don't need any fancy identity's like $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$ or some linear algebra stuff. 
