<s>[INST] Determinant of $\left(\begin{smallmatrix} -2a &a+b &a+c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{smallmatrix}\right)$ Evaluate $$D=\begin{vmatrix}
-2a &a+b  &a+c \\ 
 b+a& -2b &b+c \\ 
 c+a&c+b  & -2c
\end{vmatrix}$$
My try:
Applying $R_1 \to R_1+R_2$ we get
$$D=\begin{vmatrix}
b-a&a-b  &a+b+2c \\ 
 b+a& -2b &b+c \\ 
 c+a&c+b  & -2c
\end{vmatrix}$$
Now apply $$C_1 \to C_1+C_2$$
$$D=\begin{vmatrix}
0&a-b  &a+b+2c \\ 
 a-b& -2b &b+c \\ 
 2c+a+b&c+b  & -2c
\end{vmatrix}$$
Now apply $C_2 \to C_2 +C_3$
$$D= \begin{vmatrix}
0&2a+2c  &a+b+2c \\ 
 a-b& c-b &b+c \\ 
 2c+a+b&b-c  & -2c
\end{vmatrix}$$
Now use $R_3 \to R_3+R_2$
$$D= \begin{vmatrix}
0&2a+2c  &a+b+2c \\ 
 a-b& c-b &b+c \\ 
 2c+2a&0  & b-c
\end{vmatrix}$$
any way to proceed here using elementary operations?
 [/INST] Let $p(a,b,c) $ be the determinant. Note that each term of $p$ has degree 3 (sum of degrees of $a,b,c$).
Note that $p(a,-a,c) = 0$, hence $a+b$ divides $p$.
Similarly we see that $a+c, b+c$ divide $p$.
Hence $p$ has the form $p(a,b,c) = k (a+b)(b+c)(a+c)$ for some constant $k$.
Compute the determinant for $a=b=c={1\over 2}$ to get $k = 4$.
</s>