Determinant equal to zero : elements are angles Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle. 
I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$ 
We have the following: 
\begin{align*}&\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=\cos \beta \cdot \det \begin{pmatrix}  -1 & \cos\alpha \\  \cos\gamma & \cos\beta\end{pmatrix}-\cos \alpha \cdot \det \begin{pmatrix} \cos\gamma & \cos\alpha \\ -1 & \cos\beta\end{pmatrix}+(-1)\cdot \det \begin{pmatrix} \cos\gamma & -1  \\ -1 & \cos\gamma \end{pmatrix} \\ & = \cos \beta \cdot \left (-\cos \beta-\cos\gamma\cdot \cos\alpha\right )-\cos \alpha \cdot \left (\cos\gamma\cdot \cos\beta-(-1)\cdot \cos\alpha\right )-\left (\cos^2 \gamma -(-1)^2\right ) \\ & =-\cos^2 \beta-\cos\gamma\cdot \cos\alpha\cdot \cos \beta-\cos\alpha\cdot \cos\gamma\cdot \cos\beta- \cos^2\alpha-\cos^2 \gamma +1 \\ & =1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma \end{align*} 
Do we use here the Law of cosines?
\begin{align*}&c^2=a^2+b^2-2ab\cdot \cos \gamma \Rightarrow  \cos \gamma=\frac{a^2+b^2-c^2}{2ab} \\ &b^2=a^2+c^2-2ac\cdot \cos \beta \Rightarrow \cos \beta=\frac{a^2+c^2-b^2}{2ac} \\ &a^2=b^2+c^2-2bc\cdot \cos \alpha \Rightarrow \cos\alpha=\frac{b^2+c^2-a^2}{2bc}\end{align*}  
Or how could we continue? 
 A: If a point $P$ has trilinear coordinates $[\cos B,\cos A,-1]$ it lies on the height from vertex $C$, since the circumcenter and the orthocenter are isogonal conjugates and the distances of the circumcenter from the $BC$ and $AC$ sides are exactly given by $R\cos A$ and $R\cos B$. The claim turns out to be equivalent to the collinearity of three points at infinity, or, by duality, to the concurrency of the heights of a triangle. In a simpler way,
$$ \begin{pmatrix}c & b & a\end{pmatrix}\begin{pmatrix}\cos B & \cos A & -1 \\ \cos C & -1 & \cos A \\ -1 & \cos C & \cos B\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
hence the determinant of the given matrix is trivially zero.
A: the product of the three cosines is given by $$-\frac{\left(a^2-b^2-c^2\right)
   \left(a^2+b^2-c^2\right) \left(a^2-b^2+c^2\right)}{2
   a^2 b^2 c^2}$$
for your second sum we get
$$2 \left(\frac{\left(a^2+b^2-c^2\right)^2}{4 a^2
   b^2}+\frac{\left(a^2-b^2+c^2\right)^2}{4 b^2
   c^2}+\frac{\left(-a^2+b^2+c^2\right)^2}{4 b^2
   c^2}\right)=\frac{2 a^6-4 a^4 b^2+a^4 c^2+2 a^2 b^4+2 a^2 b^2
   c^2+b^4 c^2-2 b^2 c^4+c^6}{2 a^2 b^2 c^2}$$you must only multiply out the first sum
