2. Given that the cubic polynomial $x^{3}-a x^{2}-1003 x$ $+b$ can be divided by $x^{2}-999 x+1994$. Then the value of $b-6 a$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9
2. $\mathbf{A}$

Since $x^{2}-999 x+1994=(x-2)(x-997)$, when $x=2$ and $x=997$, by the assumption, if $x^{3}-a x^{2}-1003 x$ $+b$ can be divided by $x^{2}-999 x+1994$, then we have
$$
\left\{\begin{array}{l}
2^{3}-a \times 2^{2}-1003 \times 2+b=0, \\
997^{3}-a \times 997^{2}-1003 \times 997+b=0 .
\end{array}\right.
$$

Solving these equations yields $a=996, b=5982$. Therefore, $b-6 a=6$.