Question

Using the property of determinants and without expanding, prove that

$\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|=0$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}1& bc& a(b+c)\\ 1& ca& b(c+a)\\ 1& ab& c(a+b)\end{array}\right|$
$=\left|\begin{array}{ccc}1& bc& ab+ac\\ 1& ca& bc+ab\\ 1& ab& ca+bc\end{array}\right|$
Aplying $R_1=R_1-R_2\text{\&}{R}_2=R_2-R_3$
$\Delta =\left|\begin{array}{ccc}0& bc-ca& ac-bc\\ 0& ca-ab& ab-ca\\ 1& ab& c(a+b)\end{array}\right|$
Taking $(-1$ common from $C_2$ we get,
$\Delta =\left|\begin{array}{ccc}0& ac-bc& ac-bc\\ 0& ab-ca& ab-ca\\ 1& -ab& c(a+b)\end{array}\right|=0$
Since $R_1$ and $R_2$ are same.