TITLE: Irreducibility test in a number field
QUESTION [10 upvotes]: While thinking about this question I noticed that if $\alpha$ is a root of $f(x) = x^9 + 3 x^6 + 165 x^3 + 1$ then $$\left(\frac{\alpha^3+1}{3 \alpha}\right)^3 = -6$$ so $\mathbb{Q}(\alpha)$ contains a cube root of $6$. This shows that $3 \mid [\mathbb{Q}(\alpha) : \mathbb{Q}]$. 
Moreover, $f(x)$ factors over $\mathbb{Q}(\zeta, \sqrt[3]6)$ as
$$f(x) = (x^3 + 3 \sqrt[3]6\,x + 1)(x^3 + 3 \zeta \sqrt[3]6\,x + 1)(x^3 + 3 \zeta^{-1} \sqrt[3]6\,x + 1)$$
where $\zeta$ is a primitive third root of unity. Is there a "simple" way to use these facts to show that in fact $f(x)$ must be irreducible over $\mathbb{Q}$, e.g. by relating this to some finite field that contains a cube root of $6$?

REPLY [3 votes]: This answer elaborates on the observations made by @JyrkiLahtonen in the comments. Let $\alpha$ be a root of $x^9+3x^6+165x^3+1$ then $$\tag1 (\alpha^3+1)^3+6(3\alpha)^3=0$$ and $$\tag2 3(\alpha^3+7)^3+2(\alpha^3-8)^3=0.$$
Equation $(1)$ is already mentioned in the question and $(2)$ was found by observing that $$(1-b)(\alpha^3+a)^3+(a-1)(\alpha^3+b)^3$$ is linear in $\alpha^3$ for any $a,b$ and then picking suitable values.  From this it follows that $\mathbb{Q}(\alpha)$ contains both a cube root of $3$ and a cube root of $2$. As mentioned in the comments, this suffices to show that $\alpha$ has degree $9$ over $\mathbb{Q}$.