4. 103 Real numbers $\alpha, \beta$ satisfy the system of equations
$$\left\{\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-17=0 \\
\beta^{3}-3 \beta^{2}+5 \beta+11=0
\end{array}\right.$$

Find $\alpha+\beta$.
[Solution] The system of equations is transformed into
$$\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-14=0 \\
(\beta-1)^{3}+2(\beta-1)+14=0
\end{array}$$

Adding the two equations, we get $(\alpha-1)^{3}+(\beta-1)^{3}+2(\alpha+\beta-2)=0$,
which simplifies to $(\alpha+\beta-2)\left[(\alpha-1)^{2}+(\beta-1)^{2}-(\alpha-1)(\beta-1)+2\right]=0$.
However, when $x, y$ are both real numbers, the equation
$$x^{2}+y^{2}-x y+2=0$$

does not hold, otherwise $x=\frac{y \pm \sqrt{-3 y^{2}-8}}{2}$ would contradict the fact that $x$ is real.
Therefore, $\alpha+\beta-2=0$,
which means $\alpha+\beta=2$.