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# The value of the determinant $\begin{vmatrix}1 &a&a^2-bc\\1&b&b^2-ca\\1 &c&c^2-ab\end{vmatrix}$ is

(A) 1

(B) 0

(C) 2

(D) 3

Operating $R_1\rightarrow R_1-R_2$
$R_2\rightarrow R_2-R_3$
$\begin{vmatrix}0 &a-b&(a-b)(a+b+c)\\0&b-c&(b-c)(a+b+c)\\1 &c&c^2-ab\end{vmatrix}$
$\Rightarrow (a-b)(b-c)\begin{vmatrix}0 &1 &a+b+c\\0 &1&a+b+c\\1&c&c^2-ab\end{vmatrix}=0$
Because $R_1$ and $R_2$ are identical.
Hence (b) is the correct answer.
edited Jun 17, 2014