16. Given $\left(x^{2}+a x-3\right)(x+1)=x^{3}+b x^{2}+c x-3$, where $a, b, c$ are constants, then $b-c=(\quad)$.
(A) -4 .
(B) -3 .
(C) 1 .
(D) 4 .
Answer: D.
Solution: Expand the left side of the equation, we get $x^{3}+(a+1) x^{2}+(a-3) x-3$,
compare the coefficient of the quadratic term on the right side, we get
$$
a+1=b,
$$

compare the coefficient of the linear term, we get
$$
a-3=c,
$$

so
$$
b-c=(a+1)-(a-3)=4 \text {. }
$$

Therefore, the answer is D.