Question

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

Answer 1

Multiply $C_1,C_2$ and $C_3$ by $a,b,c$ respectively and divide by $abc$
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}{a}^2& b^2-bc& c^2+cb\\ a^2+ac& b^2& c^2-ac\\ a^2-ab& a^2+ab& c^2\end{array}\right|$
Apply $C_1+C_2+C_3$ and take out $a^2+b^2+c^2$
$\Delta =\frac{a^2+b^2+c^2}{abc}\left|\begin{array}{ccc}1& b^2-bc& c^2+cb\\ 1& b^2& c^2-ac\\ 1& b^2+ab& c^2\end{array}\right|$
Take $b$ and $c$ common from $R_2$ and $R_3$
$=\frac{a^2+b^2+c^2}{abc}.bc\left|\begin{array}{ccc}1& b-b& c+b\\ 1& b& c-a\\ 1& b+a& c\end{array}\right|$
Apply $R_2-R_1$ and $R_3-R_1$
$\Delta =\frac{a^2+b^2+c^2}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 0& b& -a-b\\ 0& a+c& -b\end{array}\right|$
$=\frac{a^2+b^2+c^2}{a}[-bc+(a+c)(a+b)]$
$=\frac{a^2+b^2+c^2}{a}[-bc+a^2+ac+ab+bc]$
$=\frac{a^2+b^2+c^2}{a}.a(a+b+c)$
$=(a+b+c)(a^2+b^2+c^2)$.