Question

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

Answer 1

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