Question

Prove the following :
$\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& {(a+b)}^2\end{array}\right|=2abc{(a+b+c)}^3.$

Answer 1

Apply $C_2-C_1$ and $C_3-C_1$.

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

Take out $(a+b+c)$ common from each of $C_2$ and $C_3$

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

Apply $R_1-(R_2+R_3)$. Then

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

Apply $C_2+\frac{1}{b}{C}_1,C_{3.}+\frac{1}{c}{C}_1$ to make two zeros in $R_1$

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

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

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

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