<s>[INST] Irreducibilty of polynomial $x^9-6x^6+282x^3-8$ over $\mathbb {Q} $ While trying to deal with the final parts of this answer I found that one needs to establish $$a=\sqrt[3]{1+\sqrt{11}}\notin\mathbb{Q} (\sqrt{11},\sqrt[3]{10})=F$$ Since both $a, F$ are of degree $6$ over $\mathbb {Q} $  the tower theorem does not help much.
Then I reasoned via contradiction. If $a\in F$ then $b=\sqrt[3]{1-\sqrt{11}}=-\sqrt[3]{10}/a\in F$ and hence $c=a+b\in F$.
One can check that $$c^3=a^3+b^3+3abc=2-3\sqrt[3]{10}c$$ or $$(c^3-2)^3=-270c^3$$ or $c$ is a root of $$f(x) =x^9-6x^6+282x^3-8$$ I checked using pari/gp that the polynomial $f(x) $ is irreducible over $\mathbb {Q} $. Further it can be observed that $f(289)$ is prime (also checked via pari/gp) so that the polynomial $f(x) $ is irreducible by Murty's criterion.
This shows that $c$ is of degree $9$ over $ \mathbb {Q} $ and hence $c\notin F$.
It can be observed that $f(x) $ can not be handled by Eisenstein as $2$ is the only prime which divides all non-leading coefficients and $4\mid 8$. I also checked the reducibility mod $3$ and the polynomial is reducible mod 3. So even that approach does not work. I don't know if using reducibility modulo other primes would help.
Is there is any other simpler way to prove the irreducibilty of $f(x) $ using hand computation?
Update: We have $$f(2x)=8(64x^9-48x^6+282x^3-1)=8g(x)$$ and it is somewhat easier to apply Murty's criterion on $g(x) $ with $g(8)=8577496063$ being prime. But I still find this unsuitable for hand calculation.
 [/INST] The polynomial $P(x)$ factors $\!\!\!\mod \!\!13$ as $(x^3+6) ( x^6+ x^3 + 3) $  (link).
Assume that $P(x)$  is reducible. We conclude that $P(x)$ is a product of two  polynomials $P_1(x)$, $P_2(x)\in \mathbb{Z}[x]$.  with $P_1(x)\equiv x^3 + 6 \!\!\!\mod \!\!13$, $P_2(x)\equiv x^6 + x^3 + 3 \!\!\!\mod\!\! 13$.
Now, $P(1) =269$, a prime number. We conclude $P_1(1)=\pm 1$, or $P_2(1)=\pm 1$.  But $P_1(1)\equiv  7 \!\!\!\mod \!\!13$, while $P_2(1)\equiv 5 \!\!\!\mod \!\!13$, contradiction.
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$\bf{Added:}$ About the factorization $\mod 13$. Our polynomial is $P(x) = Q(x^3)$, with $Q(x)= x^3 - 6 x^2 + 282 x - 8$. Now $Q(x)$ has root $7$ $\mod 13$, so we have
$Q(x) = (x-7)(x^2 + x + 3)$, and so
$P(x) = (x^3 - 7)(x^6+x^3 + 3)$. Now $x^3-7$ has no root $\mod 13$, because $7^{\frac{12}{3}}= 7^4 \not \equiv 1 \mod 13$.  Also, $x^2 + x + 3 = x^2 + 2 \cdot 7 x + 7^2 + 3 - 7^2= (x+7)^2 - 7$, and $7$ is not a quadratic residue $\mod 13$, so $x^2 + x + 3$ is irreducible.
Based on Jyrki's idea: Note that since $13\equiv 1 \mod 3$, there exists $\omega \in \mathbb{F}_{13}$, such that $\omega^3 = 1$, and $\omega\ne 1$ ( a primitive root of $1).  \mod 13$). Now, consider the action of the group of roots of $1$ on monic polynomials $f(x) \mapsto \frac{1}{\omega^{\deg f}} f(\omega x)$. The polynomials in $x^3$ are invariated by the action ( the fixed points are $x^m \cdot $ pol in $x^3$). Therefore, the a polynomial in $x^3$, its irreducible factors will be either polynomials in $x^3$, or group in orbits of size $3$ under this action. Now, if the polynomial $x^6 + x^3 + 3$ were reducible, it would equal $R(x) R(\omega x) R(\omega^2 x)$, where $R$ is a polynomial of degree $2$. But that would imply that the free term $3$ is a cube. However, we have $3^{\frac{12}{3}} = 3^4 =3 \mod 13$. Therefore, $x^6 + x^3 + 3$ is irreducible $\mod 13$.
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