9. (10th "Hope Cup" Invitational Competition Question) Let $f(x)=x^{3}-3 x^{2}+6 x-6$, and $f(a)=1, f(b)=$ -5, then $a+b=$ ( $\quad$.
A. -2
B. 0
C. 1
D. 2
9. D. Reason: Consider the monotonically increasing odd function $g(t)=t^{3}+3 t$ on $\mathbf{R}$, let $g(a-1)=3, g(b-1)=-3$, which gives $a^{3}-3 a^{2}+6 a-5=0, b^{3}-3 b^{2}+6 b+1=0$. From $g(a-1)=g(1-b)$, we get $a-1=1-b$, hence $a+b=2$.