9. Let $\alpha, \beta$ satisfy the equations respectively
$$
\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-4=0, \\
\beta^{3}-3 \beta^{2}+5 \beta-2=0
\end{array}
$$

then $\alpha+\beta=$ $\qquad$
9. 2 .

We have
$$
\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-1=0, \\
(1-\beta)^{3}+2(1-\beta)-1=0,
\end{array}
$$

which means $\alpha-1$ and $1-\beta$ are solutions to the equation $x^{3}+2 x-1=0$.
Since $x^{3}+2 x-1=0$ has only one solution, then
$$
\alpha-1=1-\beta \Rightarrow \alpha+\beta=2
$$