Obtaining Determinant without Expanding The question is #653 from Golan's Linear Algebra Every Graduate Student Should Know and while it doesn't explicitly say not to just expand and factor it, I think that's the spirit of the question.
$$
\begin{vmatrix}
-2a & a+b & a+c\\ 
a+b & -2b & b+c \\ 
c+a & c+b & -2c \\
\end{vmatrix}
$$
Since the answer is
$$
4(a+b)(b+c)(a+c)
$$
I am inclined to think it has something to do with wisely dividing out (a+b), etc. from particular rows or some other linear combination tricks but I can't seem to quite figure it out. Any ideas?
 A: 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  $F(a,b,c) = 4(a+b)(b+c)(c+a)$
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.
A: Ah, I only now realized the question was asked before. To atone for my repost sin, I'll give an answer that I hadn't seen around and seems to me to be likely what the problem aims at (inspired by the substitution proposed in the other answers), please do check that it is correct:
Using the $(a+b)=x,(b+c)=y,(a+c)=z$ substitution, we have
$$
\begin{vmatrix}
y-x-z & x & z\\ 
x & z-x-y & y \\ 
z & y & x-y-z \\
\end{vmatrix}
$$
Now $R1=R1+R2+R3$:
$$
\begin{vmatrix}
y & z & x\\ 
x & z-x-y & y \\ 
z & y & x-y-z \\
\end{vmatrix}
$$
Now you multiply column 1 by $xz$, column 2 by $xy$ and column 3 by $yz$, in order to make $xyz$ across the row. To compensate you put $\frac{1}{(xyz)^2}$ in front:
$$\frac{1}{(xyz)^2} \begin{vmatrix}
xyz & xyz & xyz\\ 
x^2z & xy(z-x-y) & y^2z \\ 
z^2x & xy^2 & yz(x-y-z) \\
\end{vmatrix}$$
Now you take out $xyz$ from the first row and do $C2=C2-C1$, $C3=C3-C1$, and you get something like: 
$$\frac{1}{(xyz)} \begin{vmatrix}
1 & 0 & 0\\ 
... & ... & ... \\ 
... & ... & ... \\
\end{vmatrix}$$
Where $...$ stands in for polynomials I don't really want to write out, but you can see only one 2x2 determinant survives, which when expanded should yield the desired $4x^2y^2z^2$, which then divides by $xyz$ to get the result $4xyz$. 
