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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.

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    $\begingroup$ If $b+c\sqrt d$ is a cube in $\Bbb Q(\sqrt d)$.... $\endgroup$ Aug 11 '19 at 17:39
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    $\begingroup$ Have you learned any field theory? This problem designed to be solved with tools of field theory. $\endgroup$ Aug 11 '19 at 17:39
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    $\begingroup$ 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. $\endgroup$ Aug 11 '19 at 17:39
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    $\begingroup$ @SimplyBeautifulArt How does that ensure integer coefficients... $\endgroup$ Aug 11 '19 at 17:40
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    $\begingroup$ @LordSharktheUnknown I'm pretty sure OP is looking for integral polynomials. $\endgroup$ Aug 11 '19 at 17:40
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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 $\sqrt[6]{2}$?

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

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