5

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.
  • 5,857
  • 2
    $x=\sqrt[3]{2}+\sqrt[3]{3} \implies x^3=5+3\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})=5+3\sqrt[3]{6}x$. Now you only have one root to care about... – chubakueno Jan 17 '14 at 01:13
  • 1
    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
  • 2
    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
  • 1
    The calculations presented by @chubakueno in the first comment should have been written as an answer ($11$ years ago!!). – User Jun 10 '25 at 02:07

4 Answers4

5

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
  • 313
  • 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
2

Try to get rid of the radicals by cubing.

$x=2^{1/3}+3^{1/3}\implies$

$x^3=2+3\cdot2^{2/3}\cdot3^{1/3}+3\cdot2^{1/3} \cdot3^{2/3}+3=5+3\cdot2^{2/3}\cdot3^{1/3}+3\cdot2^{1/3}\cdot3^{2/3}\implies$

$x^3-5=3\cdot2^{2/3}\cdot3^{1/3}+3\cdot2^{1/3}\cdot3^{2/3}\implies$

$(x^3-5)^3=$

$27\cdot2^2\cdot3+3\cdot9\cdot2\cdot2^{1/3}\cdot3^{2/3}\cdot3\cdot2^{1/3}\cdot3^{2/3}+3\cdot3\cdot2^{2/3}\cdot3^{1/3}\cdot9\cdot2^{2/3}\cdot3\cdot3^{1/3}+27\cdot2\cdot9$

$=27\cdot2\cdot3\cdot(2+3)+27\cdot2\cdot3\cdot2^{2/3}\cdot3\cdot3^{1/3}+27\cdot2\cdot3\cdot2^{1/3}\cdot3\cdot3^{2/3}$

$=27\cdot2\cdot3\cdot x^3=162x^3,$

so $(x-5)^3=x^9-15x^6+75x^3-125=162x^3$; i.e., $x^9-15x^6-87x^3-125=0.$

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
2

This is a comment not an answer:

Here is a slight modification of J.W. Tanner's nice idea to yield a fast path to their polynomial:

$$ x = \sqrt[3]{2} + \sqrt[3]{3} \rightarrow \\x^3 = 2 + 3 \sqrt[3]{3*2^2} + 3\sqrt[3]{3^2*2} + 3 \rightarrow \\ x^3-5 = 3 \sqrt[3]{2*3} \left(\sqrt[3]{2} + \sqrt[3]{3} \right) \rightarrow \\ x^3 - 5 = 3x \sqrt[3]{6} \rightarrow \\(x^3 -5)^3 = 27*6 x^3 $$

And from here you can expand and combine like terms to recover their result.

It is essentially an identical derivation but by not necessarily expanding everything out it's much easier to see structurally what's going on.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Sidharth Ghoshal
  • 18,359
2

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
  • 12,824