$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$. 
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \},$$ but I'm stuck proving that it's closed under inverse. If that is true, then the minimal polynomial should be of 3rd degree. True or not, I can't find any such polynomial.
Solution: As @José Carlos Santos mentioned below:
$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha,$$
from which it follows
$$(x - \sqrt[3]3-\sqrt[3]9)(x^2 + x(\sqrt[3]3+\sqrt[3]9) + (\sqrt[3]3+\sqrt[3]9)^2) - 9(x - \sqrt[3]3 - \sqrt[3]9)= x^3 - 12 - 9(\sqrt[3]3+\sqrt[3]9) -9x + 9(\sqrt[3]3+\sqrt[3]9) = x^3 - 9x - 12 = p(x);~~ p(\alpha) = 0.$$

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 $$(3+\alpha)^3=3(1+\alpha)^3\iff 12+9\alpha-\alpha^3=0$$
