Find the minimal polynomial over $Q$ of the root $(\sqrt[2]{3})/(1+\sqrt[3]{2})$. I'm new to field theory so please help!
(I'm new to the platform so excuse my sloppy notation) So far, I set a= $(\sqrt[2]{3})/(1+\sqrt[3]{2})$, and by squaring both sides, I got $a^2+2^{4/3}a^2+2^{2/3}a^2=3$. I tried manipulating this expression quite a bit, but I can't seem to be able to get rid of the fractional exponents. I tried isolating the cubic root to one side and cubing both sides first as well and no luck. Am I supposed to be using other principles?
Setting $a = \frac{3^{1/2}}{1+2^{1/3}}$, as is already pointed out in the OP's question, squaring gives $$ \tag{$\dagger$} a^2 = \frac{3}{1+2.2^{1/3} +2^{2/3}} \implies a^2(1+2^{2/3}+2^{4/3}) = 3. $$ One way of proceeding from here is to notice that if we let $\beta = 2^{2/3}$ then $$ 1+2^{2/3} + 2^{4/3} = 1+\beta+ \beta^2 =\frac{\beta^3-1}{\beta-1}= \frac{3}{\beta-1} $$ Thus substituting this back into ($\dagger$) we find $a^2 = \beta-1$. But $\beta$ obviously satisfies $g(y)=t^3-4$, hence we deduce that $$ (a^2+1)^3-4 = 0 \iff a^6+3a^4+3a^2-3=0 $$ Thus $a$ satisfies $f(t)= t^6+3t^4+3t^2-3\in \mathbb Z[t]$. One still needs to check whether this is the minimal polynomial or not of course.