8. Given real numbers $\alpha, \beta$ satisfy $\alpha^{3}-3 \alpha^{2}+5 \alpha=1, \beta^{3}-3 \beta^{2}+5 \beta=5$, then $\alpha+\beta=$
8. 2

From the conditions, we can deduce that
$$
(\alpha-1)^{2}+2(\alpha-1)+2=0, \quad(1-\beta)^{3}+2(1-\beta)+2=0 .
$$

The function $f(x)=x^{3}+2 x+2$ is monotonically increasing on $R$, therefore $\alpha-1=1-\beta$, so $\alpha+\beta=2$.