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
Each entry of the third column can be written using the first two columns:
\( bc(b+c) = bc(a+b+c) - abc \)
\( ca(c+a) = ca(a+b+c) - abc \)
\( ab(a+b) = ab(a+b+c) - abc \)
So \(C_3 = (a+b+c)\,C_2 - abc\,C_1\). The column operation \(C_3 \to C_3 - (a+b+c)\,C_2 + abc\,C_1\) does not change the value of the determinant, and it makes the third column zero:
\( \begin{vmatrix}1 & bc & bc(b+c)\\1 & ca & ca(c+a)\\1 & ab & ab(a+b)\end{vmatrix} = \begin{vmatrix}1 & bc & 0\\1 & ca & 0\\1 & ab & 0\end{vmatrix} \)
A determinant with a column of zeros is 0.