Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
 $$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?

Yet another argument: It is easy to check that
\begin{equation}
\left(\begin{array}{ccc}
b+c & a & a \\ b & c+a & b \\ c & c & a+b
\end{array}\right)
= L N ,
\end{equation}
where the matrices $L$ and $N$ are defined by
\begin{equation}
L = \left(\begin{array}{ccc}
0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0
\end{array}\right)
\qquad \text{and} \qquad
N = \left(\begin{array}{ccc}
0 & c & b \\ c & 0 & a \\ b & a & 0
\end{array}\right)
.
\end{equation}
Since the determinant is multiplicative, it thus suffices to prove that $\det L = 2$ and $\det N = 2abc$. This is easy.
