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I want to calculate the minimal polynomial of $\alpha=\sqrt{2}+\sqrt[3]{2}$ over $\mathbb{Q}$. Usually these kind of problems can be solved by doing stuff like $(\alpha-\sqrt{2})^2=\sqrt[3]{4}\Leftrightarrow\dots$ and so on until you get a polynomial equation in $\alpha$ and coefficients in $\mathbb{Q}$ to 0. But in this specific case I fail to find it, although I can argue for it to be of degree 6 (moreover my computer says it's $x^6-6x^4-4x^3+12x^2-24x-4$). For example, doing the above I end up at $\alpha^3-3\sqrt{2}\alpha^2+6\alpha+2\sqrt{2}-2=0$ where squaring doesn't get rid of all the squareroots.

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If you have $$\alpha^3-3\sqrt2\alpha^2+6\alpha-2\sqrt2-2=0$$ rearrange as $$\alpha^3+6\alpha-2=\sqrt2(3\alpha^2+2)$$ and square. This gets rid of the $\sqrt2$.

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  • $\begingroup$ Oh wow, sometimes I'm bilnd. Thanks a lot! $\endgroup$ – user424862 May 1 '17 at 8:33

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