Finding the degree of $\sqrt[3]{2} + \sqrt[3]{3}$ over $\mathbb Q$ I am practicing writing down random algebraic numbers and finding their degrees over $\mathbb Q$ and have fumbled when coming to $\sqrt[3]{2} + \sqrt[3]{3}$. Mathematica tells me the minimal polynomial is $x^9-27 x^6-135 x^3-729$, which is a polynomial in $x^3$. I figured by cubing the element I could show that the cube has degree 3, which is some partial progress, but I have failed to even do that. If anyone has any neat tricks up their sleeves about these kinds of results, I would be forever indebted to you.

First, Mathematica is wrong. The minimal polynomial is $-125-87X^3-15X^6+X^9$.
The irreducible equation for $\sqrt[3]2$ over $\Bbb Q$ is $X^3-2$, and the irreducible equation for $\sqrt[3]2+\sqrt[3]3$ over $\Bbb Q(\sqrt[3]3\,)$would apparently be $F(X)=(X-\sqrt[3]3\,)^3-2=X^3-3\sqrt[3]3X^2+3\sqrt[3]9X-5$. Now, to get a $\Bbb Q$-polunomial out of $F$, you need to multiply it by its two conjugates, gotten by (1) replacing $\sqrt[3]3$ by $\omega\sqrt[3]3$, and (2) replacing $\sqrt[3]3$ by $\omega^2\sqrt[3]3$, where $\omega=\frac12(-1+\sqrt{-3}\,)$ is a chosen primitive cube root of $1$. The conjugate polynomials are
\begin{align}
F_1(X)&=X^3-3\omega\sqrt[3]3X^2+3\omega^2\sqrt[3]9X-5\>,\>\text{and}\\
F_2(X)&=X^3-3\omega^2\sqrt[3]3X^2+3\omega\sqrt[3]9X-5\,.
\end{align}
I’ll leave it to you to do the multiplication $FF_1F_2$. It’s tedious by hand, but far easier than computing a nine-by-nine determinant!
