Question

Show that :

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

Answer 1

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

$=\left|\begin{array}{ccc}bc-a^2& ca–b^2& ab{-}^2\\ ab+bc+ca-2a^2& ab+bc+ca-2b^2& ab+bc+ca-2c^2\\ (a+b)(a+c)& (b+c)(b+a)& (c+a)(c+b)\end{array}\right|[R_2\to R_2+2R_1]$

$=3\left|\begin{array}{ccc}bc-a^2& ca–b^2& ab{-}^2\\ ab+bc+ca-2a^2& ab+bc+ca-2b^2& ab+bc+ca-2c^2\\ a^2& b^2& c^2\end{array}\right|[R_3\to R_3-R_2]$

$=3(ab+bc+ca)\left|\begin{array}{ccc}bc& ca& ab\\ 1& 1& 1\\ a^2& b^2& c^2\end{array}\right|[R_1\to R_1+R_3;R_2\to R_2+2R_3]$

$=3(ab+bc+ca)(b-a)(c-a)\left|\begin{array}{ccc}bc& -c& -b\\ 1& 0& 0\\ a^2& b+a& c+a\end{array}\right|[C_2\to C_2-C_1;C_3\to C_3-C_1]$

$=3(ab+bc+ca)(b-a)(c-a)\{-c(c+a)+b(b+a\left)\right\}$

$=3(ab+bc+ca)(b-a)(c-a)(-c^2-ac+b^2+ab)$

$=3(b-c)(c-a)(a-b)(a+b+c)(bc+ca+ab)$ $\left[henceproved\right]$