Question

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

Answer 1

$\begin{array}{c}\left|\begin{array}{ccc}{\left(b+c\right)}^2& a^2& a^2\\ b^2& {\left(c+a\right)}^2& b^2\\ c^2& c^2& {\left(a+b\right)}^2\end{array}\right|\\ ={\left(b+c\right)}^2\left[{\left(c+a\right)}^2\times {\left(a+b\right)}^2-b^2\times c^2\right]-\\ a^2\left[b^2\times {\left(a+b\right)}^2-b^2\times c^2\right]+a^2\left[b^2{c}^2-c^2{\left(c+a\right)}^2\right]\\ =b^2+c^2+2ab\left[\left(c^2+a^2+2ac\right)\times \left(a^2+b^2+2ab\right)-b^2{c}^2\right]-a^2\\ \left[b^2\times \left(a^2+b^2+2ab\right)-b^2{c}^2\right]+a^2\left[b^2{c}^2-c^2\left(c^2+a^2+2ac\right)\right]\\ onsolvingandtakingcomonweget\\ =2abc{\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}^2\\ =2abc{\left(a+b+c\right)}^2\\ proved\end{array}$