Is the polynomial $f(x) = x^9 - 9x^6 - 27x^3 - 27$ irreducible over $\mathbb{Q}?$ I think it is because of Eisenstein's applied to the prime $3.$
Is it the minimal polynomial of $x = e^{2 \pi i/3} + \sqrt[3]{2}$ over $\mathbb{Q}$? Wolfram Alpha says the answer is $x^6+3 x^5+6 x^4+3 x^3+9 x+9.$
I am a bit confused.
Note that Eisenstein's criterion requires $p^2$ to not divide the constant term. Your polynomial is reducible. Note that the minimal polynomial of $\omega$ is the quadratic $x^2 + x + 1$, so the extension $\Bbb Q(\omega, \sqrt[3]{2})$ is of degree six, and thus $\omega + \sqrt[3]{2}$ has a minimal polynomial of degree dividing six.
(Wolfram alpha factors $ x^9 - 9x^6 - 27x^3 - 27$ as $x^3-3x^2+3x-3$ times the degree six polynomial you cite in your question.)