Is $x^{14}+x^7+1$ irreducible over $Q[x]$? Is $x^{14}+x^7+1$ irreducible over $\mathbb{Q}[x]$?
I think it is, but I'm not able to justify it using any existing criterion (e.g. Eisenstein). Any help?
Indeed, this is a question I encounter on a linear algebra one. The original question gives a $8 \times 8$ real matrix satisfying $A^{21}=I$ and asks to prove that $\mathbb{R}^8$ can be decomposed into the direct sum of 4 2-dimensional vector subspace invariant w.r.t. $A$. 
My attempt was to find the minimal polynomial then discuss several cases, which need the factor of $x^{21}-1$. Any hint on the original question is also appreciated. 

$x^{14}+x^7+1=(x^7-\omega)(x^7-\omega^2)$
$$=(x^7-\omega^7)(x^7-\omega^{14}) ----(1)$$
Factorizing each factors further we get 
$$(x^7-\omega^7)=(x-\omega)(x^6+x^5\omega +x^4\omega^2 +x^3 +x^2\omega +x \omega^2+1)$$
$$=((1+x^3+x^6)+(x^2+x^5)\omega +(x+x^4)\omega^2 ---(2)$$
Similarly $(x^7-\omega^{14})=(x-\omega^2)((1+x^3+x^6)+(x^2+x^5)\omega^2+(x+x^4)\omega) -----(3)$
$$let\,p(x)=(1+x^3+x^6)\,,\,q(X)=(x^2+x^5)\,,\,r(x)=(x+x^4)$$
Using (1) , (2) , and (3) we get
$$x^{14}+x^7+1=(x-\omega)(x-\omega^2)(p(x)+q(x)\omega +r(x)\omega^2)(p(x)+q(x)\omega^2 +r(x)\omega)$$
$$=(1+x+x^2)((p(x))^2+(q(x))^2+(r(x))^2-p(x)q(x)-q(x)r(x)-r(x)p(x))$$
