Minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$

Another way to do this is to interpret $\alpha$ as a $\mathbb{Q}$-linear map from $\mathbb{Q}(\sqrt[3]{2})$ to itself. With respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, this is represented by the matrix $$\begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}.$$ The characteristic and minimal polynomial is $$t^3 - (0+0+0) t^2 + 3 \cdot \mathrm{det} \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix} t - \mathrm{det} \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix}$$ $$= t^3 - 6t - 6.$$