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I'm trying to calculate the minimal polynomial $f^\alpha_\mathbb{Q}$ for $$ \alpha = \sqrt[3]{2} + \sqrt[3]{4}. $$ But I don't know how to do this. Does anyone have a hint? I tried to square $\alpha$, then I get $\alpha^2 = \sqrt[3]{4} + 4 + \sqrt[3]{16}$. But I'm not sure where that's going.

Does anyone know a structured method for solving these kind of problems, instead of just trying some things manually?

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    $\begingroup$ $$\alpha^3=2+4+6\alpha$$ $\endgroup$ – lab bhattacharjee Feb 10 '18 at 11:55
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Let $t=\sqrt [3] 2$ so that $t^3=2$ and $a=t+t^2$

Every polynomial expression in $t$ can be reduced to a quadratic in $t$ using $t^3=2$. Three such expressions will enable $t$ to be eliminated.

$a^2=t^4+2t^3+t^2=2t+4+t^2$

$a^3=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2=\text{ (we notice immediately) } 6a+6$

Had that easy observation not been available, standard elimination of $t$ and $t^2$ from $a, a^2, a^3$ using your favourite (linear) method would leave at most a cubic.

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Using the Binomial Theorem $$\alpha^3=(\sqrt[3]{2} + \sqrt[3]{4})^3=2+3\sqrt[3]{16}+3\sqrt[3]{32}+4=6+3\sqrt[3]{8}(\sqrt[3]{2}+\sqrt[3]{4})=6(1+\alpha)$$ so the required polynomial is $$\alpha^3-6\alpha-6$$ Your best bet when dealing with expressions containing $n$th roots is to raise the expression to the power $n$ and hope for a nice simplification.

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    $\begingroup$ Great thanks! I hadn't thought at all about Binomial theorem but that seems to simplify it a big deal! Lastly, to show that this is irreducible we can say that it is Eisenstein for $p = 3$ right? $\endgroup$ – Sigurd Feb 10 '18 at 12:03
  • $\begingroup$ Yes! This example works perfectly due to $6=2\cdot3$. $\endgroup$ – TheSimpliFire Feb 10 '18 at 12:04
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I suppose that you know the formula $(a+b)^3=a^3+b^3+3ab(a+b)$

Hence by cubing both sides we have $$\alpha^3= 2+4+3*2*(\sqrt[3] {2}+\sqrt[3] {4})$$ but we have $\sqrt[3] {2}+\sqrt[3] {4}=\alpha$ hence we get $$\alpha^3= 2+4+6\alpha$$ $$\Rightarrow \alpha^3-6\alpha-6=0$$

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