The value of the determinant $\begin{vmatrix}b+c &a-b &a\\c+a &b-c &b\\a+b &c-a &c\end{vmatrix}$, is
Answer & explanation
Correct answer: option 2
We have,
$\begin{vmatrix}b+c &a-b &a\\c+a &b-c &b\\a+b &c-a &c\end{vmatrix}$
$=\begin{vmatrix}a+b+c&-b &a\\b+c+a &-c &b\\c+a+b &-a &c\end{vmatrix}$ [Applying $C_1→C_1 + C_3; C_2 →-C_2-C_3$]
$=-(a+b+c)\begin{vmatrix}1&b &a\\1&c &b\\1&a &c\end{vmatrix}$
$=-(a+b+c) \begin{vmatrix}1&b &a\\0&c-b &b-a\\0&a-b &c-a\end{vmatrix}$ [Applying
$R_2→R_2-R_1, R_3 → R_3-R_1$]
$=-(a+b+c)(a^2 + b^2 + c^2 -ab-bc - ca)$
$=-(a^3+b^3+c^3-3abc)$