Obtaining Determinant without Expanding

Consider $F(a,b,c) = \begin{vmatrix} -2a & a+b & a+c\\ a+b & -2b & b+c \\ c+a & c+b & -2c \\ \end{vmatrix}$

With some work we can show that $F(a,b,c)$ is a cyclic symmetric polynomial of degree $3$.

Now, we can use the properties of such polynomials to evaluate the determinant. When $a=-b$ the determinant becomes $0$. Hence $(a+b)$ is a factor. Similarly $(b+c)$, $(c+a)$ are factors.

Hence $F(a,b,c) = k(a+b)(b+c)(c+a)$. Now, to determine $k$ set $a=1, b=1, c=0$. We get $k=4$.

Hence $F(a,b,c) = 4(a+b)(b+c)(c+a)$

I don't see any obvious tricks, but if you put $(x,y,z)=(a+b,\,b+c,\,c+a)$ and rewrite the matrix as $$ A=\pmatrix{ y-x-z &x &z\\ x &z-x-y &y\\ z &y &x-y-z}, $$ then $$ PAP=B=\pmatrix{ 0&2x&2z\\ 2x&0&2y\\ 2z&2y&0},\ \text{ where }P=\pmatrix{ 0&1&1\\ 1&0&1\\ 1&1&0}. $$ Hence $\det(A)=\det(B)/\det(P)^2$. You don't need to expand $\det(P)$ to calculate its value. In fact, since $P=ee^T-I$ (here $e$ denotes the all-one vector), we immediately get $\det(P)=e^Te-1=2$. However, to calculate $\det(B)$, I cannot think of a better method than Sarrus' rule.