Question

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

Answer 1

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

Apply $R_1\to R_1+R_2+R_3$

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

factor out $a+b+c$ from first row

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

Apply $C_1\to C_1-C_2,C_2\to C_2-C_3$

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

Factor out $a+b+c$ from first and second column

$=(a+b+c{)}^3\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& 2b\\ 0& 1& c-a-b\end{array}\right|$

Expand along first row

$=(a+b+c{)}^3[1-0]=(a+b+c{)}^3$