Question

Prove the following :
$\left|\begin{array}{ccc}b+c& c+a& a+b\\ a+b& b+c& c+a\\ c+a& a+b& b+c\end{array}\right|=2\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

Answer 1

Operating $C_1+(C_2+C_3)$ and taking out $2$ from newly formed $C_1$, we get
$△=2\left|\begin{array}{ccc}a+b+c& c+a& a+b\\ a+b+c& b+c& c+a\\ a+b+c& a+b& b+c\end{array}\right|$
Apply $C_2-C_1$ and $C_3-C_1$
$\therefore △=2\left|\begin{array}{ccc}a+b+c& -b& -c\\ a+b+c& -a& -b\\ a+b+c& -c& -a\end{array}\right|$
$=2{(-1)}^2\left|\begin{array}{ccc}a+b+c& b& c\\ a+b+c& a& b\\ a+b+c& c& a\end{array}\right|$
$=2\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$ applied $C_1-(C_2+C_3)$