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Find the minimal polynomial

  1. Nov 12, 2008 #1

    Office_Shredder

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    1. The problem statement, all variables and given/known data

    Find the minimal polynomial of [tex]\frac{\sqrt{3}}{1+2^{1/3}}[/tex] over Q

    we'll call this x

    2. Relevant equations
    I wish I knew some :(

    3. The attempt at a solution


    By taking powers of x, I was able to show that the extension Q(x) has degree six (since 21/3 and sqrt(3) are both independently inside Q(x)) and hence the minimal polynomial has degree six. So then I took a general degree six polynomial, plugged in x, and got six equations in six unknowns. This is less than elegant, and I ended with a system of equations (luckily three of the six unknown coefficients were zero):

    255b + 261d + 595f = -171
    150b + 222d + 460f = -144
    105b + 159d + 375f = -108

    solving for b,d,f. An attempt at a numerical solution doesn't convince me this has a rational solution, but I wouldn't be surprised if there was a computational error preceding this. Is there a better way to do this?
     
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

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    How about starting from the equation x=sqrt(31)+2^(1/3)? So (x-sqrt(31))^3=2. Expand the left side and move all of the terms involving sqrt(31) to one side and everything else to the other. Now square.
     
  4. Nov 12, 2008 #3

    Office_Shredder

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    Sorry, you posted before I fixed the latex. Different value for x

    EDIT: Same principle worked though. Thanks a ton
     
  5. Nov 12, 2008 #4

    Dick

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    Fast, aren't I? Same idea. x=sqrt(3)/(1+2^(1/3)) -> x2^(1/3)=sqrt(3)-x. Cube, rearrange and then square.
     
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