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