I want to calculate the degree $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]$.
It holds that $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}][\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]\cdot 2$$ and $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$ since we know that the polynomial $f=x^4-2x^2-1\in \mathbb{Q}[x]$ is irreducible, and $\rho\in \mathbb{C}$ is a root of $f$.
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has $A=\pmatrix{0&3\\3&0}$. It has three faces, so the rank of $G$ is $\chi(G)=2$. The reciprocal of Ihara's $\zeta$ function can be evaluated
$$
\frac{1}{\zeta_G(u)}={(1-u^2)^{\chi(G)-1}\det(I - Au + (k-1)u^2I)}\\
=(1-u^2) (4u^4-5u^2+1)
$$
Looking at the orientation of the edges...
