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Mathematics
Problem
Prove that
∣
−
2
a
a
+
b
a
+
c
b
+
a
−
2
b
b
+
c
c
+
a
c
+
b
−
2
c
∣
=
4
(
b
+
c
)
(
c
+
a
)
(
a
+
b
)
\begin{vmatrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{vmatrix} = 4(b+c)(c+a)(a+b)
−
2
a
b
+
a
c
+
a
a
+
b
−
2
b
c
+
b
a
+
c
b
+
c
−
2
c
=
4
(
b
+
c
)
(
c
+
a
)
(
a
+
b
)
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