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. D.
$$
\begin{array}{l}
\text { Given } f(x)=x^{3}+3 x^{2}+6 x+14 \\
=(x+1)^{3}+3(x+1)+10,
\end{array}
$$

Let $g(y)=y^{3}+3 y$. Then $g(y)$ is an odd function and strictly increasing.
And $f(a)=(a+1)^{3}+3(a+1)+10=1$,
$$
f(b)=(b+1)^{3}+3(b+1)+10=19 \text {, }
$$

Then $g(a+1)=-9, g(b+1)=9$,
$$
g(-b-1)=-9 \text {. }
$$

Thus, $g(a+1)=g(-b-1)$, which means $a+1=$ $-b-1$. Therefore, $a+b=-2$.