Find minimal polynomial of element Find minimal polynomial of element $3-2\sqrt[3]{2}-\sqrt[3]{4}$ $\in {\displaystyle \mathbb {Q}(\sqrt[3]{2}) } $ over the field $ {\displaystyle \mathbb {Q} } $.
Thank you for any help. 
 A: Here's a systematic way to find the minimal polynomial, which works in general setting. We know that $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is a basis of $\mathbb{Q}(\sqrt[3]{2})$ as a vector space over $\mathbb{Q}$. Now consider the following relations:
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot1 = 3-2\sqrt[3]{2}-\sqrt[3]{4}$$
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{2} = -2 + 3\sqrt[3]{2} - 2\sqrt[3]{4}$$
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot \sqrt[3]{4} = -4 - 2\sqrt[3]{2} + 3 \sqrt[3]{4}$$ 
These three equations can be written in matrix notation in the following way:
$$ (3-2\sqrt[3]{2}-\sqrt[3]{4})\left[
\begin{array}{c}
  1\\
  \sqrt[3]{2}\\
  \sqrt[3]{4}
\end{array}
\right]  =  \left[
\begin{array}{ccc}
  3&-2&-1\\
  -2&3&-2\\
  -4&-2&3
\end{array}
\right]\left[
\begin{array}{c}
  1\\
  \sqrt[3]{2}\\
  \sqrt[3]{4}
\end{array}
\right] $$
This gives us that $(3-2\sqrt[3]{2}-\sqrt[3]{4})$ is an eigenvalue of the matrix on the right. Hence it's a zero of the polynomial $\det(xI-M)$. Obviously as the degree of it is $3$ and $(3-2\sqrt[3]{2}-\sqrt[3]{4}) \not \in \mathbb{Q}$ we get that it must be the minimal polynomial of $(3-2\sqrt[3]{2}-\sqrt[3]{4})$. Evaluating the determinant isn't a hard thing to do and you will get that the minimal polynomial is:
$$f(x) = x^3 - 9x^2 + 15x + 29$$
