I've been searching around stack exchange trying to find a good example on this but most of them seem to require less calculation than this one seems to take.
Find the minimal polynomial of $1+\sqrt[3]2+\sqrt[3]4$ over $\mathbb Q$.
The way I've been approaching it is $$x=1+\sqrt[3]2+\sqrt[3]4\Rightarrow x-1-\sqrt[3]2=\sqrt[3]4$$ then cubing both sides and working from there but it seems to take an awful lot of calculation. Is there a quicker way? I also tried to reverse engineer an answer by checking it on Wolfram Alpha but I think they gave me the wrong answer ($x^3-3x^2-3x-1$) because it wasn't working out for me. But maybe I just did the calculations wrong.
First note that $$1+\sqrt[3]2+\sqrt[3]4=\frac{2-1}{\sqrt[3]2-1}=\frac1{\sqrt[3]2-1}$$ The minimal polynomial of $\sqrt[3]2$ is $x^3-2$. That of $\sqrt[3]2-1$ can be found by substituting $x\to x+1$: $(x+1)^3-2=x^3+3x^2+3x-1$. The minimal polynomial of the original expression is then obtained by reversing the coefficients (to effect the reciprocal): $-x^3+3x^2+3x+1$ or $x^3-3x^2-3x-1$. Wolfram Alpha is right on this one.