SOLUTION — $\left|\begin{array}{ccc}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|$
$=\left|\begin{array}{ccc}0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & c^2-a b\end{array}\right|$
Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b+c\\0 & 1 & a+b+c \\1 & c & c^2-a b\end{array}\right|=0$
( $\because$ Rows $R_1$ and $R_2$ are identical)
So, The correct option is (A).