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?

You can use the rule of Sarrus in this case:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
\begin{align}
&=(b+c)(c+a)(a+b) +abc +abc 
- c(c+a)a - cb(b+c) -ab(a+b)\\
&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2 +2abc -c^2a-a^2c-c^2b-b^2c -a^2b-b^2a\\
&=4abc.
\end{align}
