What is the minimal polynomial of $\alpha = \frac{3^{1/2}}{1+2^{1/3}} $ over $\mathbb{Q}$? 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.
 A: 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

A: 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.
