Polynomial in $\mathbb{Q}[x]$ with root $\sqrt[3]{2}+\sqrt[3]{3}$ What is a polynomial $P(x)\in \mathbb{Q}[x]$ with root $\sqrt[3]{2}+\sqrt[3]{3}$? 
I write $x=\sqrt[3]{2}+\sqrt[3]{3}$, so $(x-\sqrt[3]{2})^3=3$, but the expansion of the left side contains two cube roots again. What can we do?
 A: Expand $[(x- \sqrt[3]{2})^3 - 3][(x- j\sqrt[3]{2})^3 - 3][(x- j^2\sqrt[3]{2})^3 - 3] = 0$, where $j$ is a third root of unity in $\mathbf{C}$.
A: A tedious but elementary answer:
Let $x = \sqrt[3]{2} + \sqrt[3]{3}$. Clearly, any power of $x$ can be obtained from $\sqrt[3]{2}$ and $\sqrt[3]{3}$ by addition and multiplication. Thus, $x^k$ can be written in the form: $x^n = \sum_{k=0}^2\sum_{k=0}^2 a^{(n)}_{i,j} \sqrt[3]{2}^k \sqrt[3]{3}^l$. The values $\sqrt[3]{2}^k \sqrt[3]{3}^l$ are rationally independent, but there are only $9$ of these.
Thus, if you were to write out at least $10$ distinct powers of $x$, there is guaranteed to be a linear relation between them. This relation will give you a polynomial with root $x$. This is not something you want to do by hand, but it certainly can be done in principle and on a computer. If you notice that $x^3$ is particularly nice, it is fairly natural to try and write out $x^6,x^9$, and luckily a linear relation will already appear at this stage.
The explicit answer is $x^9-15 x^6-87 x^3-125 = 0$, but I admit I took a look at Wolfram's output rather than compute this myself.
