8 Let the function $f(x)=x^{3}+3 x^{2}+6 x+14$, and $f(a)=1, f(b)=19$, then $a$ $b=(\quad)$.
(A) 2
(B) 1
(C) 0
(D) -2
8 From
$$
\begin{aligned}
f(x) & =x^{3}+3 x^{2}+6 x+14 \\
& =(x+1)^{3}+3(x+1)+10,
\end{aligned}
$$

let $g(y)=y^{3}+3 y$, then $g(y)$ is an odd function and monotonically increasing. And
$$
\begin{array}{l}
f(a)=(a+1)^{3}+3(a+1)+10=1, \\
f(b)=(b+1)^{3}+3(b+1)+10=19,
\end{array}
$$

so $g(a+1)=-9, g(b+1)=9, g(-b-1)=-9$, thus
$$
g(a+1)=g(-b-1),
$$

which means
$$
a+1=-b-1,
$$

hence $a+b=-2$. Therefore, the answer is D.