Question

Prove that

$\left|\begin{array}{ccc}bc-a^2& ca-b^2& ab-c^2\\ -bc+ca+ab& bc-ca+ab& bc+ca-ab\\ (a+b)(a+c)& (b+c)(b+a)& (c+a)(c+b)\end{array}\right|=3.(b-c)(c-a)(a-b)(a+b+c)(ab+bc+ca)$

Answer 1

Consider, $\left|\begin{array}{ccc}bc-a^2& ca-b^2& ab-c^2\\ -bc+ca+ab& bc-ca+ab& bc+ca-ab\\ (a+b)(a+c)& (b+c)(b+a)& (c+a)(c+b)\end{array}\right|$
$=\left|\begin{array}{ccc}bc-a^2& ca-b^2& ab-c^2\\ -bc+ca+ab& bc-ca+ab& bc+ca-ab\\ a^2+ab+bc+ac& b^2+ab+bc+ac& c^2+ab+bc+ac\end{array}\right|$
$R_3\to R_3+R_1$
$=\left|\begin{array}{ccc}bc-a^2& ca-b^2& ab-c^2\\ -bc+ca+ab& bc-ca+ab& bc+ca-ab\\ ab+2bc+ac& ab+bc+2ac& 2ab+bc+ac\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}(b-a)(a+b+c)& (c-b)(a+b+c)& ab-c^2\\ 2c(a-b)& 2a(b-c)& bc+ca-ab\\ c(b-a)& a(c-b)& 2ab+bc+ac\end{array}\right|$
$=(a-b)(b-c)\left|\begin{array}{ccc}-(a+b+c)& -(a+b+c)& ab-c^2\\ 2c& 2a& bc+ca-ab\\ -c& -a& 2ab+bc+ac\end{array}\right|$
$R_2\to R_2+{2R}_3$
$=(a-b)(b-c)\left|\begin{array}{ccc}-(a+b+c)& -(a+b+c)& ab-c^2\\ 0& 0& 3bc+3ca+3ab\\ -c& -a& 2ab+bc+ac\end{array}\right|$
Expanding along second row, we get
$=3(a-b)(b-c)(ab+bc+ca)(c-a)(a+b+c)$