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

(A) Let $\Delta = \left|\begin{array}{lll}1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b)\end{array}\right|$
Apply the column operation $C_{3} \rightarrow C_{3} + C_{2}$:
$\Delta = \left|\begin{array}{lll}1 & bc & ab+bc+ca \\ 1 & ca & ab+bc+ca \\ 1 & ab & ab+bc+ca\end{array}\right|$
Taking $(ab+bc+ca)$ as a common factor from $C_{3}$:
$\Delta = (ab+bc+ca) \left|\begin{array}{lll}1 & bc & 1 \\ 1 & ca & 1 \\ 1 & ab & 1\end{array}\right|$
Since columns $C_{1}$ and $C_{3}$ are identical,the value of the determinant is $0$.
$\therefore \Delta = (ab+bc+ca) \times 0 = 0$.