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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?

Mika H.
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    $x=\sqrt[3]{2}+\sqrt[3]{3} \implies x^3=5+3\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})=5+3\sqrt[3]{6}x$. Now you only have one root to care about... – chubakueno Jan 17 '14 at 01:13
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    http://www.wolframalpha.com/input/?i=minimal+poly+%282%29%5E%281%2F3%29%2B3%5E%281%2F3%29 – David P Jan 17 '14 at 01:14
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    I described a general procedure here. You need to adapt my example a little bit. The example involves $\sqrt2+\sqrt3$ and produces a set of linear equations over four variables that you solve to find the coefficients of a 4th-degree monic polynomial. For your example you'll need to solve a set of nine linear equations. – MJD Jan 17 '14 at 01:31
  • @MJD I actually find your approach very useful for (primitive) CAS-coding purposes, but let there be a space for creativity today :) – chubakueno Jan 17 '14 at 01:35

2 Answers2

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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 $\mathbf{C}$.

user121926
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  • Could you please explain why this should be in $\mathbb{Q}[x]$? – Mika H. Jan 17 '14 at 01:23
  • The coefficients will be symmetric polynomials over $\mathbf{Q}$ of the numbers $\sqrt[3]{2}$, $\sqrt[3]{2}j$, $\sqrt[3]{2}j^2$, and can therefore be expressed in terms of the elementary symmetric polynomials of these three numbers. But the values of these elementary symmetric polynomials are the coefficients of the polynomial $x^3 - 2$, whose roots are the three numbers. – user121926 Jan 17 '14 at 01:36
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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.

Jakub Konieczny
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