Question

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

Answer 1

$\left|\begin{array}{ccc}{(b+c)}^2& ab& ca\\ ab& {(a+c)}^2& bc\\ ac& bc& {(a+b)}^2\end{array}\right|=\frac{1}{abc}\left|\begin{array}{ccc}a{(b+c)}^2& a^{2}b& {ca}^2\\ {ab}^2& b{(a+c)}^2& b^{2}c\\ {ac}^2& {bc}^2& c{(a+b)}^2\end{array}\right|$ [Multiplying $R_1$ by $a,R_2$ by $b,R_3$ by $c$]

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

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

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

$=\frac{{(a+b+c)}^2}{ab}.ab.2ab\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ 0& 0& 1\end{array}\right|[C_1\to C_1+C_3{C}_2\to C_2+C_3]$

$=2ab(a+b+c{)}^1[(b+c)(c+a)-ab]$

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

$=2ab(a+b+c{)}^2(b+c+a)$

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