Determinant of matrices without expanding Show that $$\begin{array}{|ccc|}
-2a & a + b & c + a \\
a + b & -2b & b + c \\
c + a & c + b & -2c
\end{array} = 4(a+b)(b+c)(c+a)\text{.}$$
I added the all rows but couldn't get it.
 A: Let $x=b+c,y=c+a,z=a+b$. We claim that
$$
\left|\begin{pmatrix}
x-y-z & z & y\\
z & y-z-x & x\\
y & x & z-x-y
\end{pmatrix}\right|=4xyz.
$$
When $x=0$, add column 1 to columns 2 and 3 to obtain
$$
\left|\begin{pmatrix}
-y-z & -y & -z\\
z & y & z\\
y & y & z
\end{pmatrix}\right|=0.
$$
Thus by symmetry, $xyz$ divides the determinant. Setting $x=y=z=1$ yields
$$
\left|\begin{pmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{pmatrix}\right|=
\left|\begin{pmatrix}
-1 & 0 & 0\\
1 & 0 & 2\\
1 & 2 & 0
\end{pmatrix}\right|=4.
$$
Therefore the determinant is $4xyz$, since it is of degree 3.
A: We can deduce from the structure of the matrix that (a) the determinant will be a symmetric polynomial in $a,b,c$ with every monomial with non-zero coefficient having degree $3$, and (b) that the coefficient of $a^3$, $b^3$, and $c^3$ will be $0$.  So
$$
 \det= k\,(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+m\,abc
$$
for some constants $k,m$.  To find $k,m$, we substitute in values of $(a,b,c)$ into the matrix.


*

*When $a=0$, $b=1$, $c=1$ we get $8=\begin{vmatrix} 0 & 1 & 1 \\ 1 & -2 & 2 \\ 1 & 2 & -2 \\ \end{vmatrix}=\det=2k$, so $k=4$.

*When $a=1$, $b=-1$, $c=1$ we get $0=\begin{vmatrix} -2 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & -2 \\ \end{vmatrix}=\det=4 \times 2-m$, so $m=8$.
Thus
\begin{align*}
 \det &= 4(a^2 b+ a^2 c+ b^2 a+ b^2 c+ c^2 a+ c^2 b)+8abc \\
 &= 4(a+b)(b+c)(c+a)
\end{align*}
as desired.
