1. Given that $a$, $b$, and $c$ are constants, and for any real number $x$ we have
$$
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \text {. }
$$

then $a b c=$ . $\qquad$
$$
\text { Two, 1. }-9 \text {. }
$$

From the problem, we know
$$
\begin{array}{c}
x^{3}+2 x+c=(x+1)\left(x^{2}+a x+b\right) \\
=x^{3}+(a+1) x^{2}+(a+b) x+b .
\end{array}
$$

Thus, $a+1=0, a+b=2, b=c$
$$
\begin{array}{l}
\Rightarrow a=-1, b=c=3 \\
\Rightarrow a b c=-9
\end{array}
$$