Question

Using the property of determinants and without expanding.
$\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|=0$

Answer 1

Let $A=\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|=\left|\begin{array}{ccc}1& bc& ab+ac+bc\\ 1& ca& bc+ba+ca\\ 1& ab& ca+cb+ab\end{array}\right|$ (using $C_3\to C_3+C_2$)

Taking $ab+bc+ca$ common from $C_3,$ we get
$=(ab+bc+ca)\left|\begin{array}{ccc}1& bc& 1\\ 1& ca& 1\\ 1& ab& 1\end{array}\right|=(ab+bc+ca)\times 0=0$
[Since, the two columns $C_1$ and $C_3$ are identical]