Question

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

Answer 1

$L:H:S=\left|\begin{array}{ccc}{(b+c)}^2& ab& ca\\ ab& {(c+a)}^2& bc\\ ca& bc& {(a+b)}^2\end{array}\right|$

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

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

$R_1+C_1-C_3,C_2-C_2-C_3$

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

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

$R_3+R_3-(R_1+R_2)$

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

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

$\frac{(a+b+c{)}^2}{ab}\left|\begin{array}{ccc}a+b+c& a^2& a^2\\ b^2& bc+ab& b^2\\ U& U& 2ab\end{array}\right|$

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

$R:H:S$ (Proved)