<s>[INST] 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?
 [/INST] 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.
</s>