Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$ 
i have to prove:
  $$\begin{vmatrix}
-2a &a+b  &a+c \\ 
 b+a&-2b  &b+c \\ 
 c+a&c+b  &-2c 
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$

I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?
 A: First of all, note that the determinant is going to be a polynomial in a, b, and c of degree 3. Secondly, note that the polynomial is cyclic.
Substitute $a = -b$, to get,
$$\begin{vmatrix}
-2a &0  &c+a \\ 
 0&2a  &c-a \\ 
 c+a&c-a  &-2c 
\end{vmatrix} = -2a(-4ac - (c-a)^2 + (c+a)^2) = 0$$
This shows that $a+b$ is factor of the polynomial.
Since $a+b$ is a factor and the polynomial is cyclic, $b+c, c+a$ are also factors. And since the degree of the polynomial is 3, there is at most only a constant factor left. Let the constant factor be $k$.
Thereby,
$$\begin{vmatrix}
-2a & a+b  &c+a \\ 
 a+b& -2b  &b+c \\ 
 c+a& b+c  &-2c 
\end{vmatrix} = k(a+b)(b+c)(c+a)$$
Put $a = 0, b = 1, c = 1$ to get,
$$\begin{vmatrix}
0 & 1  & 1 \\ 
 1& -2  & 2 \\ 
 1& 2  &-2 
\end{vmatrix} = k(1)(2)(1) \Rightarrow k = 4$$ 
Thus,
$$\begin{vmatrix}
-2a & a+b  &c+a \\ 
 a+b& -2b  &b+c \\ 
 c+a& b+c  &-2c 
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
A: Brute force: expand using Cramer's rule:
$$
\begin{split}
\det A = &-2a\left( (-2b)(-2c) - (b+c)^2 \right)\\
         &-(b+a)\left( (-2a)(-2c) - (a+c)^2\right)\\
         &+(c+a)\left( (a+b)(b+c)-(-2b)(a+c)\right)
\end{split}
$$
and simplify out.
Similarly multiply out the RHS. It is painful, but maybe less painful than looking for the clever transformations.
