Question

Show that $\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|=2(a+b+c{)}^3$

Answer 1

$\left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|$

$\left|\begin{array}{ccc}a+b+2c& a& b\\ 0& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|$

$=\left|\begin{array}{ccc}a+b+c& 0& -c-a-b\\ 0& a+b+c& -(a+b+c)\\ c& a& c+a+2b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& 0& -1\\ 0& (a+b+c)& -(a+b+c)\\ c& a& c+a+2b\end{array}\right|$

$=(a+b+c{)}^2\left|\begin{array}{ccc}1& 0& -1\\ 0& 1& -1\\ c& a& c+a+2b\end{array}\right|$

$=(a+b+c{)}^2\int 1(a+c+2b+a)+c\left(1\right)$

$=2(a+b+c{)}^3$

Hence proved.