Find the minimal polynomial with real root

Daveyboy
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Find the minimal polynomial with root 21/3 + 21/2.

I would just use maple but I do not have it installed on this machine.
I found the polynomial and verified that this is indeed a root. I only have Eisenstiens criterion for determining whether it is irreducible, and I can not apply it in this case. Do you have another method? I have not tried substituting x=x+1 or x=x-1 or other substitutions.

The polynomial is x6-6x4-4x3+12x2-24x-4
 
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Try computing [ \mathbb{Q}(\sqrt[3]{2} + \sqrt{2}) : \mathbb{Q} ].

Or... the dimension of the vector space spanned by powers of \sqrt[3]{2} + \sqrt{2}.

Or... something else that would tell you information about the minimal polynomial.
 
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Daveyboy said:
Find the minimal polynomial with root 21/3 + 21/2.

I would just use maple but I do not have it installed on this machine.
I found the polynomial and verified that this is indeed a root. I only have Eisenstiens criterion for determining whether it is irreducible, and I can not apply it in this case. Do you have another method? I have not tried substituting x=x+1 or x=x-1 or other substitutions.

The polynomial is x6-6x4-4x3+12x2-24x-4

You don't need to show the polynomial is irreducible, do you? You just want to show it has no rational roots. Look at the rational root test. If that has a rational root, the root must divide 4.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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