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I've tried to solve this by algebraic manipulation: putting the relation in equation form, raising it to powers, rearranging terms, rewriting some of them in terms of $\alpha$ and reading off the minimal polynomial from the final equation, written only with rational coefficients. It didn't work. But I think that's the way to go about it as I haven't studied splitting fields yet and so I don't expect the solution to require them.

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  • $\begingroup$ Set $x=\alpha$ and try to get rid of the radicals. $\endgroup$ – Brandon Thomas Van Over Oct 20 '15 at 9:29
  • $\begingroup$ That's what I've tried. $\endgroup$ – guta1 Oct 20 '15 at 9:31
  • $\begingroup$ And what did you obtain? $\endgroup$ – Brandon Thomas Van Over Oct 20 '15 at 9:33
  • $\begingroup$ $ 2^{1/3} (2^{1/3} +2) \alpha^2 = 3 - \alpha^2 $ and from here I rewrote the term $(2^{1/3} +2) \alpha^2$ as $3^{1/2} \alpha$ $\endgroup$ – guta1 Oct 20 '15 at 9:42
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    $\begingroup$ My rule of thumb to simplify complicated expression is "get rid of the most ugly piece first". $$\begin{align}\frac{\sqrt{3}}{\alpha} - 1 = 2^{1/3} \implies & (\sqrt{3}-\alpha)^3 = 2\alpha^3\\ & \iff 3\sqrt{3} - 3\cdot 3\alpha + 3\sqrt{3}\alpha^2 - \alpha^3 = 2\alpha^3\\ & \iff \sqrt{3}(1+\alpha^2) = (\alpha^2+3)\alpha\\ \implies & (\alpha^2+3)^2\alpha^2 - 3(1+\alpha^2)^2 = 0\\ & \iff \alpha^6+3\alpha^4+3\alpha^2-3 = 0\end{align} $$ $\endgroup$ – achille hui Oct 20 '15 at 9:45
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Although it's a bit "unfair" and probably not allowed in an examination I can not resist showing how easy such a calculation can be done with Groebner bases and (for example) Macaulay 2: $x$ stands for $3^{1/2}$, $y$ stands for $2^{1/3}$ and $z$ for $\alpha$. All the relations are encoded in the ideal id1 and the elimination is done with a standard trick with the map phi:

i1 : R=QQ[x,y,z]

o1 = R

o1 : PolynomialRing

i2 : i1=ideal(x^2-3,y^3-2,z*(1+y)-x)

             2       3
o2 = ideal (x  - 3, y  - 2, y*z - x + z)

o2 : Ideal of R

i3 : S=QQ[z]

o3 = S

o3 : PolynomialRing

i5 : phi=map(R/i1,S)

                       R
o5 = map(-----------------------------,S,{z})
           2       3
         (x  - 3, y  - 2, y*z - x + z)

                           R
o5 : RingMap ----------------------------- <--- S
               2       3
             (x  - 3, y  - 2, y*z - x + z)

i6 : ker phi

            6     4     2
o6 = ideal(z  + 3z  + 3z  - 3)

o6 : Ideal of S
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Since, $\alpha=\frac{3^{1/2}}{1+2^{1/3}}=3^{-1/2}(1-2^{1/3}+2^{2/3})$, let $\beta=\alpha^2=2^{2/3}-1$. Then $$ \begin{bmatrix} \beta^0\\\beta^1\\\beta^2\\\beta^3 \end{bmatrix} = \left[\begin{array}{r} 1&0&0\\-1&0&1\\1&2&-2\\3&-6&3 \end{array}\right] \begin{bmatrix} 1\\2^{1/3}\\2^{2/3} \end{bmatrix} $$ With a bit of linear algebra, or perhaps inspection, it is not too difficult to see that $$\beta^3+3\beta^2+3\beta-3=0$$ Therefore, $$\alpha^6+3\alpha^4+3\alpha^2-3=0$$ which is irreducible by Eisenstein's Criterion.

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  • $\begingroup$ Ok, I see... But does that mean the answer given by achille hui is wrong ? The two polynomials clearly differ, and I've gone through the calculations for the first answer and they seem correct. Am I missing something ? $\endgroup$ – guta1 Oct 20 '15 at 11:33
  • $\begingroup$ @guta1: no, I had an error, which has been fixed. $\endgroup$ – robjohn Oct 20 '15 at 11:56

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