 Solution

Eisenstein criterion says that a primitive integer polynomial is irreducible over $\mathbb{Q}$ if some prime divides all coefficients except the leading one, does not divide the leading coefficient, and its square does not divide the constant coefficient. The passage from primitive integer factorization to rational factorization is Gauss lemma for polynomials.

Translate the variable:This is Eisenstein at two. Translation is an automorphism of $\mathbb{Q}[x]$, so $x^8+1$ itself is irreducible.

Set $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. Since $\zeta$ is a root of that irreducible degree-eight polynomial, $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Also $c=(\zeta +{\zeta }^{-1})/2$, so $\mathbb{Q}(c)\subseteq \mathbb{Q}(\zeta )$ and $\zeta$ solves $T^2-2cT+1=0$. The extension degree is at most two. It is not one, since every element of $\mathbb{Q}(c)$ is real and $\zeta$ is not. The tower law for field extensions now gives