10. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial equation $x^{3}-10 x+11=0$.
(1) Given that $x_{1}, x_{2}, x_{3}$ all lie in the interval $(-5,5)$, find the integer parts of these three roots;
(2) Prove: $\arctan x_{1}+\arctan x_{2}+\arctan x_{3}=\frac{\pi}{4}$.
(1) Let $f(x)=x^{3}-10 x+11$, after calculation we get $f(-4)=-64+40+11=-13, f(1)=1-10+11=2>0, \\
f(2)=8-20+11=-1
\end{array}
$$

By the Intermediate Value Theorem, we know $x_{1} \in(-4,-3), x_{2} \in(1,2), x_{3} \in(2,3)$.
Therefore, $\left[x_{1}\right]=-4,\left[x_{2}\right]=1,\left[x_{3}\right]=2([x]$ represents the integer part of $x)$.
(2) Let $\alpha=\arctan x_{1}, \beta=\arctan x_{2}, \gamma=\arctan x_{3}$, from (1) we get $\alpha \in\left(-\frac{\pi}{2},-\frac{\pi}{4}\right)$, $\beta, \gamma \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \Rightarrow \alpha+\beta+\gamma \in\left(0, \frac{3 \pi}{4}\right)$
By the relationship between roots and coefficients, we have $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=0, \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=-10, \\ x_{1} x_{2} x_{3}=-11 .\end{array}\right.$
Thus, $\tan (\alpha+\beta)=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}, \tan (\alpha+\beta+\gamma)=\frac{\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}+x_{3}}{1-\frac{x_{1}+x_{2}}{1-x_{1} x_{2}} \cdot x_{3}}$
$$
=\frac{x_{1}+x_{2}+x_{3}-x_{1} x_{2} x_{3}}{1-x_{1} x_{2}-\left(x_{1}+x_{2}\right) \cdot x_{3}}=\frac{x_{1}+x_{2}+x_{3}-x_{1} x_{2} x_{3}}{1-\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)}=1 \text {. }
$$

In summary, $\alpha+\beta+\gamma=\arctan x_{1}+\arctan x_{2}+\arctan x_{3}=\frac{\pi}{4}$.