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$\sqrt{2} +\sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{0}=0 $

How to find this equation? I tried Cardano's method, noting that $$\sqrt{2} +\sqrt[3]{3} = \sqrt[3]{\sqrt{8}} +\sqrt[3]{3}$$

It means $$\begin{align} -\frac{q}{2}+\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=3 \\[4pt] -\frac{q}{2}-\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=\sqrt{8} \end{align}$$

but this system doesn't have natural solutions.

Maybe it has rational $p$ and $q$?

Blue
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demsp
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  • Use the method given in Example 4 on p. 3 of this old class handout of mine. – Dave L. Renfro Mar 23 '19 at 12:34
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    By the way, the context of your question is not clear. Do you simply want to know a method for finding such an algebraic equation (with integer coefficients, something you neglected to say by the way), or do you want to find a method that BEGINS with Cardano's formula? I don't think beginning with Cardano's formula will help much because one can show (with some work) that there is no cubic polynomial with integer coefficients that has $\sqrt{2} + \sqrt[3]{3}$ as a root. – Dave L. Renfro Mar 23 '19 at 12:44
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    The minimal polynomial for $\sqrt{2}+\sqrt[3]{3}$ is has degree 6, so Cardano's method of solving cubic equations won't be any help. – Blue Mar 23 '19 at 13:11
  • Thank you very much – demsp Mar 23 '19 at 20:26
  • @DaveL.Renfro Can you please update your link to "this old calss handout of mine"? The current link is unavailable. Thanks! – Eric Bolinger Jun 29 '20 at 18:09
  • @Eric Bolinger: It was archived at a Math Forum discussion group post, which for the last few years of their discussion groups (which ran from 1996 to 2017 or 2018) allowed attachments to be posted, and which was still available for a year or two after they ended the discussion groups. However, they seem to have taken all the tens (hundreds?) of thousands of math posts down a few months (a year?) ago. The handout's title is "Proving Irrationality by the Rational Roots Theorem" and it might turn up at some point on the internet somewhere, but right now I can't find a copy anywhere. – Dave L. Renfro Jun 29 '20 at 19:15
  • @Eric Bolinger: For a more interesting example than any on that handout, see this answer of mine. Some of the same ideas I tried to emphasize in the handout are here also, such as not having to multiply everything out when you only need the leading coefficient and the constant coefficient, and non-reliance on calculators (or hand-computed decimal computations, at least when not accompanied by proved error bounds sufficient to warrant the desired claims). – Dave L. Renfro Jun 29 '20 at 19:22

2 Answers2

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Let $x=\sqrt 2 + \sqrt[3]{3}$. Then, $$ x-\sqrt 2 = \sqrt[3]{3} $$ Cubing both sides and rearranging we have, $$ x^3+6x-3=\sqrt 2 (3x^2+2) $$ Squaring both sides and rearranging we have, $$ x^6-6x^4-6x^3+12x^2-36x+1=0 $$

Awe Kumar Jha
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In the same spirit as Awe Kumar Jha, let $x=\sqrt 2 + \sqrt[3]{3}$, that is to say $$x-\sqrt 2 =\sqrt[3]{3}\implies (x-\sqrt 2) ^3=3$$ Expand and group terms to get $$x^3-3 \sqrt{2} x^2+6 x-(2 \sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.

Claude Leibovici
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