The value of $\begin{vmatrix}1&bc&bc(b+c)\\1&ca&ca(c+a)\\1&ab&ab(a+b)\end{vmatrix}$ is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
Given determinant:
$D = \begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(a+b) \\ 1 & ab & ab(a+b) \end{vmatrix}$
Subtract first row from second and third rows (R2-R1, R3-R1):
$D = \begin{vmatrix} 1 & bc & bc(b+c) \\ 0 & ca-bc & ca(a+b) - bc(b+c) \\ 0 & ab-bc & ab(a+b) - bc(b+c) \end{vmatrix}$
Expand along first column:
$D = 1 \cdot \begin{vmatrix} ca-bc & ca(a+b)-bc(b+c) \\ ab-bc & ab(a+b)-bc(b+c) \end{vmatrix}$
Compute 2×2 determinant:
$=(ca-bc)(ab(a+b)-bc(b+c)) - (ab-bc)(ca(a+b)-bc(b+c))$
Factor and simplify: $D = 0$ (after simplification, all terms cancel)