$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$

Note that$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha.$$

Let's have $q=\sqrt[3]{3}\quad$ and $\quad\alpha=q+q^2$.

$$q+q^2+q^3=q(1+q+q^2)\iff3+\alpha=q(1+\alpha)$$

Now since $q^3=3$ then $\alpha$ is solution of $$(3+\alpha)^3=3(1+\alpha)^3\iff 12+9\alpha-\alpha^3=0$$

If $\root 3 \of 3$ and $\root 3 \of 9$ are algebraic, then so is $\root 3 \of 3 + \root 3 \of 9$. The polynomial for $\root 3 \of 3$ is obviously $x^3 - 3$, and $x^3 - 9$ for $\root 3 \of 9$.

I wonder if maybe we can just add up the polynomials? We thus get $2x^3 - 12$. By the fundamental theorem of algebra, this has three solutions.

Those solutions are $\root 3 \of 6$, $\omega \root 3 \of 6$ and $\omega^2 \root 3 \of 6$, where $\omega$ is the complex cubic root of 1 with positive imaginary part. None of those numbers are equal to $\root 3 \of 3 + \root 3 \of 9$, so this seems to be a dead end.

However, it does suggest the idea of cubing $\root 3 \of 3 + \root 3 \of 9$ and seeing if that suggests a polynomial, which was probably the thought process behind Jose's answer.