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Given roots of form $$a, ~~\sqrt[3]{b+c\sqrt{d}}, \sqrt[3]{b-c\sqrt{d}}$$ $a,b,c,d\in {\mathbb Z}$,
does there exist a cubic polynomial with integer coefficients with above roots?

I tried few examples and I seem to find a cubic polynomial every time XD
Is there a proof of this?

NOTE: I came across this while trying to solve this question.

AgentS
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    If $b+c\sqrt d$ is a cube in $\Bbb Q(\sqrt d)$.... – Angina Seng Aug 11 '19 at 17:39
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    Have you learned any field theory? This problem designed to be solved with tools of field theory. – Rushabh Mehta Aug 11 '19 at 17:39
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    Short answer: yes, as these are all algebraic. Take a polynomial that has the first as a root, multiply it by a polynomial with the second as a root, and multiply by another polynomial with the third as a root. – Simply Beautiful Art Aug 11 '19 at 17:39
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    @SimplyBeautifulArt How does that ensure integer coefficients... – Rushabh Mehta Aug 11 '19 at 17:40
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    @LordSharktheUnknown I'm pretty sure OP is looking for integral polynomials. – Rushabh Mehta Aug 11 '19 at 17:40
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    Sorry, with the restriction of this being cubic the answer is no in general. But the above works to show you can construct such a polynomial. – Simply Beautiful Art Aug 11 '19 at 17:43
  • A little bit, but not too much to be able to talk with you geniuses :) If I understand correctly, if we extend rationals using $\sqrt{d}$, we can find a cube root for $b+c\sqrt{d}$. This means finding a polynomial with integer coefficients only is not possible? – AgentS Aug 11 '19 at 17:44
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    The product of the three future roots is not a rational number, thus there can't be a cubic with those three numbers as roots... – DonAntonio Aug 11 '19 at 17:57
  • Oh right the product may not always be a perfect cube unless I force $a=0$; then $x$ factors out leaving just a quadratic with cube-root coefficients.. hmm – AgentS Aug 11 '19 at 18:06
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    @DonThousand when $b+c\sqrt d$ is a cube in $\Bbb Q(\sqrt d)$ then one will have a polynomial with integer coefficients. – Angina Seng Aug 11 '19 at 18:08

1 Answers1

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No, it should be at least a 6-th degree polynomial in general.

Let $(b,c,d)=(0,1,2)$. What's the minimal polynomial for $\sqrt[6]{2}$?

Dzoooks
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    I haven't studied abstract algebra, but it seems $x^6-2$ ? I get it! Thank you so much :) – AgentS Aug 11 '19 at 17:47