Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4abc$

(A) Let $\Delta = \left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$.
Applying $R_{1} \rightarrow R_{1} - R_{2} - R_{3}$ to $\Delta$,we get:
$\Delta = \left|\begin{array}{ccc}0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b\end{array}\right|$.
Expanding along $R_{1}$,we obtain:
$\Delta = 0 \cdot \left|\begin{array}{cc}c+a & b \\ c & a+b\end{array}\right| - (-2c) \cdot \left|\begin{array}{cc}b & b \\ c & a+b\end{array}\right| + (-2b) \cdot \left|\begin{array}{cc}b & c+a \\ c & c\end{array}\right|$.
$\Delta = 2c(b(a+b) - bc) - 2b(bc - c(c+a))$.
$\Delta = 2c(ab + b^2 - bc) - 2b(bc - c^2 - ac)$.
$\Delta = 2abc + 2cb^2 - 2bc^2 - 2b^2c + 2bc^2 + 2abc$.
$\Delta = 4abc$.