<s>[INST] 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?
 [/INST] Let me give a solution without using the rule of Sarrus for fun. $\require{cancel}$
\begin{align}
& \begin{vmatrix}
b+c&a&a \\
b&c+a&b \\
c&c&a+b \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
a(b+c)&ab&ca \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&-2bc \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \tag{$R_1 \to R_1 - R_2 - R_3$} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&0 \\
ab&b(c+a)&-2ab \\
ca&bc&0 \\
\end{vmatrix} \tag{$C_3 \to C_3 - C_1 - C_2$} \\
&= \frac{1}{a\cancel{bc}}\, [-(-2\cancel{bc})]\, \begin{vmatrix}
ab&-2ab \\
ca&0 \\
\end{vmatrix} \\
&= \frac{2}{\cancel{a}} \, [-(-2ab)(c\cancel{a})] \\
&= 4abc
\end{align}
</s>