The minimal polynomial of $\alpha = e^{2\pi i/3} + \sqrt[3]{2}$ over $\mathbb{Q}$ is given by the polynomial $f(x) = x^9 - 9x^6 - 27x^3 - 27$. This polynomial has nine roots, which include $\alpha$ and other expressions of the form $e^{2r\pi i/3} + \sqrt[3]{2}e^{2s\pi i/3}$ for integers $r, s \in \{0, 1, 2\}$. The irreducibility of $f(x)$ over $\mathbb{Q}$ is suggested but requires proof.
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The numbers $e^{2r\pi i/3} + \sqrt[3]{2}e^{2s\pi i/3}$, with $r,s \in\{0,1,2\}$, all satisfy that equation. So that gives you the nine roots of $f(x)$, of which $\alpha$ is one. I'm guessing that $f(x)$ is irreducible over $\mathbb{Q}$ but I don't see how to prove that.kalish said:I started by setting $\alpha= e^{2\pi i/3} + \sqrt[3]{2}.$ Then I obtained $f(x) = x^9 - 9x^6 - 27x^3 - 27$ has $\alpha$ as a root.
How can I proceed to find the minimal polynomial of $\alpha$ over $\mathbb{Q},$ and identify its other roots?