Find the minimal polynomial

[Office_Shredder]

Homework Statement

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

we'll call this x

Homework Equations

I wish I knew some :(

The Attempt at a Solution

By taking powers of x, I was able to show that the extension Q(x) has degree six (since 2^1/3 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?

 

[Dick]

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.

 

[Office_Shredder]

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

EDIT: Same principle worked though. Thanks a ton

 

[Dick]

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