TITLE: Find a polynomial with integer coefficients
QUESTION [6 upvotes]: Find a polynomial $p$ with integer coefficients for which $a = \sqrt{2} + \sqrt[3]{2}$ is a root. That is find $p$ such that for some non-negative integer $n$, and integers $a_0$, $a_1$, $a_2$, ..., $a_n$, $p(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$, and $p(a) = 0$.
I do not know how to solve this. It is very challenging. Also, if you name any theorem please describe it in a way that is easy to understand. If you just name it, I won't be able to understand it. (My math might not be/is not as good as yours.)
Thanks for any help!

REPLY [4 votes]: Here's a general approach to this kind of problem.
[NOTE: I misread the question as involving $\sqrt{2}$ and $\sqrt[3]{a}$ instead of $\sqrt{2}$ and $\sqrt[3]{2}$, whose sum is being called $a$. This means that some of the specific algebraic calculations that follow aren't exactly the ones the original questioner needs, but the overall structure is the same whatever you're taking the cube root of so I'll leave the mistake in. You could just put $a=2$ in what I wrote to get an answer to the original question.]
Consider those roots -- the square root of 2, and the cube root of $a$. There are actually two square roots of 2 (which we usually write as $\sqrt{2}$ and $-\sqrt{2}$, but as far as equation-solving goes they're interchangeable). And there are three cube roots of $a$; if we write $\omega=(-1+i\sqrt{3})/2$ they are $\sqrt[3]{a},\omega\sqrt[3]{a},\omega^2\sqrt[3]{a}$. (The point is that $1,\omega,\omega^2$ are the things whose cube is 1, just as $1,-1$ are the things whose square is 1.)
If we have an integer polynomial with, $\sqrt{2}+\sqrt[3]{a}$ as a root, then it "can't tell the difference" between that and, say, $-\sqrt{2}+\omega^2\sqrt[3]{a}$ so that had better be a root as well. (If you want more details, the technical term to look up is "conjugate" or, more specfically, "Galois conjugate".)
Therefore, an integer polynomial with $\sqrt{2}+\sqrt[3]{a}$ as a root must have all six of these "conjugates" as roots. So we have
$$(x-(\sqrt{2}+\sqrt[3]{a}))(x-(-\sqrt{2}+\sqrt[3]{a}))(x-(\sqrt{2}+\omega\sqrt[3]{a}))(x-(-\sqrt{2}+\omega\sqrt[3]{a}))(x-(\sqrt{2}+\omega^2\sqrt[3]{a}))(x-(-\sqrt{2}+\omega^2\sqrt[3]{a}))$$
and now expanding this out gives the polynomial we need.
In practice it may be easier first to consider the cubic polynomial
$$(x-(\sqrt{2}+\sqrt[3]{a}))(x-(\sqrt{2}+\omega\sqrt[3]{a}))(x-(\sqrt{2}+\omega^2\sqrt[3]{a}))$$
which turns out, if I've done my algebra right, to equal $x^3-3\sqrt{2}x^2+6x-2\sqrt2-a$. And then we multiply this by its conjugate (with the signs of all the $\sqrt2$s flipped) to get the degree-6 polynomial we need.
So the general recipe is: take the root you need, find all its conjugates, and multiply together all the corresponding $(x-\textrm{root})$ factors. The resulting polynomial's coefficients will all come out rational.