Field extension of degree three It's well known that
$$\mathbb{Q}(\sqrt{2},\sqrt{5}) = \mathbb{Q}(\sqrt{2} + \sqrt{5})$$
This property is also true with the cubic root (for a great general theorem), but I want to prove this via an elementary proof. Anyone know how to prove that
$$[\mathbb{Q}(\sqrt[3]{2} + \sqrt[3]{5}) : \mathbb{Q}] = 9$$ ?
The minimal polynomial must be extremely large and very difficult to be computed.
 A: Let $\alpha=\sqrt[3]{3}$ and $\beta=\sqrt[3]{5}$.
Write $(\alpha+\beta)^i$ for $i=0, \ldots, 8$
in the basis 
$\{1, \alpha, \alpha^2, \beta, \alpha \beta, \alpha^2 \beta, \beta^2, \alpha \beta^2, \alpha \beta^2, \alpha^2 \beta^2\}$:
$$
\begin{pmatrix}
  1 \\ \alpha+\beta \\ (\alpha+\beta)^2 \\
  (\alpha+\beta)^3 \\ (\alpha+\beta)^4 \\ (\alpha+\beta)^5 \\
  (\alpha+\beta)^6 \\ (\alpha+\beta)^7 \\ (\alpha+\beta)^8
\end{pmatrix}
=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 0 \\
8 & 0 & 0 & 0 & 0 & 3 & 0 & 3 & 0 \\
0 & 23 & 0 & 17 & 0 & 0 & 0 & 0 & 6 \\
0 & 0 & 53 & 0 & 40 & 0 & 35 & 0 & 0 \\
334 & 0 & 0 & 0 & 0 & 93 & 0 & 75 & 0 \\
0 & 709 & 0 & 613 & 0 & 0 & 0 & 0 & 168 \\
0 & 0 & 1549 & 0 & 1322 & 0 & 1117 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
  1 \\ \alpha \\ \alpha^2 \\ \beta \\ \alpha\beta \\ \alpha^2\beta \\ \beta^2 \\ \alpha \beta^2 \\ \alpha^2 \beta^2
\end{pmatrix}
$$
Now prove that the matrix is non-singular (for instance, see here)
