$\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4 a b c$
$\Delta=\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$
$=\left|\begin{array}{ccc}0 & -2 c & -2 b \\ b & c+a & b \\ c & c & a+b\end{array}\right| \quad$ [Applying $\left.R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right)\right]$
$=\left|\begin{array}{ccc}0 & -2 c & -2 b \\ b & c+a-b & 0 \\ c & 0 & a+b-c\end{array}\right| \quad$ [Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $\left.C_{3} \rightarrow C_{3}-C_{1}\right]$
$=0\left|\begin{array}{cc}c+a-b & 0 \\ 0 & a+b-c\end{array}\right|-(-2 c)\left|\begin{array}{cc}b & 0 \\ c & a+b-c\end{array}\right|-2 b\left|\begin{array}{cc}b & c+a-b \\ c & 0\end{array}\right|$ [Expanding along $\left.R_{1}\right]$
$=2 c[b(a+b-c)-0[-2 b[0-c(c+a-b)]$
$=2 b c[a+b-c]-2 b c[b-c-a]$
$=2 b c[(a+b-c)-(b-c-a)]$
$=4 a b c$
Hence proved.