Nice identity

It depends on how much Galois Theory you know. Since $\zeta = e^{\frac{2\pi i }{13}}$ is a primitive $13$ th root of unity, the Galois group $\Gamma(\mathbb{Q}(\zeta),\mathbb{Q})$ is isomorphic to $\mathbb{Z}_{13}^\star$ to the multiplicative group of nonzero integers modulo $13$ , and hence is cyclic of order $12$ . Since this group has a unique subgroup of index $2$ , there is a unique intermediate field $\mathbb{Q} \subseteq F \subseteq \mathbb{Q}(\zeta)$ which has degree $2$ over $\mathbb{Q}$ , and this field must contain the desired number $\cos\tfrac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13}$ (the question tells us that this number satisfies a quadratic over $\mathbb{Q}$ , and it certainly belongs to $\mathbb{Q}(\zeta)$ ).

If $\alpha$ generates the cyclic group $\Gamma(\mathbb{Q}(\zeta),\mathbb{Q})$ , then $\alpha^2$ generates the subgroup of index $2$ , and so $F$ is the fixed field of $\alpha^2$ , in that $F \; = \; \big\{ z \in \mathbb{Q}(\zeta) \; \big| \; \alpha^2(z) = z\big\}$ One generator $\alpha$ is the $\mathbb{Q}$ -automorphism such that $\alpha(\zeta) = \zeta^2$ , and therefore elements of $F$ are $\begin{aligned} A & = \; \zeta + \alpha^2(\zeta) + \alpha^4(\zeta) + \alpha^6(\zeta) + \alpha^8(\zeta) + \alpha^{10}(\zeta) \\ B & = \; \alpha(\zeta) + \alpha^3(\zeta) + \alpha^5(\zeta) + \alpha^7(\zeta) + \alpha^9(\zeta) + \alpha^{11}(\zeta) \end{aligned}$ and that is what I wrote down originally.

You can use the fact that $A \; = \; \sum_{\beta \in G} \beta(\zeta) \hspace{2cm} B \; = \; \sum_{\beta \in \alpha G} \beta(\zeta)$ where $G$ is the subgroup of $\Gamma(\mathbb{Q}(\zeta),\mathbb{Q})$ of index $2$ generated by $\alpha^2$ , to show that $A$ and $B$ must be the roots of a quadratic equation.