10. (12th IMO Problem) Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following conditions:
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}$, where $i=1,2, \cdots, 1997$.
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.

Find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value.
10. Solution Let the real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the given conditions, and there exist $x_{i}$ and $x_{j}$, such that $\sqrt{3}>x_{i} \geqslant x_{j}>-\frac{1}{\sqrt{3}}$, then let $m=\frac{1}{2}\left(x_{i}+x_{j}\right), h=\frac{1}{2}\left(x_{i}-x_{j}\right)=x_{i}-m=m-x_{j}$.

Since $(m+h)^{12}+(m-h)^{12}=2 \sum_{0 \leq t=21 \leq 12} C_{12}^{t} m^{12-k} h^{k}$, and it increases as $h>0$ increases, therefore, when taking $h=\min \left\{\sqrt{3}-m, m-\left(-\frac{1}{\sqrt{3}}\right)\right\}$. And by replacing $x_{i}$ and $x_{j}$ with $x^{\prime}=m+h, x^{\prime} ;=m-h$, the sum of the variables remains unchanged, while the sum of the 12th powers of the variables increases.

Therefore, the maximum value of the sum of the 12th powers can only be achieved in the following scenario: among the variables $x_{1}, x_{2}, \cdots, x_{1997}$, at most one takes a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$, while the other variables take values of $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.
Let $u$ variables take the value $-\frac{1}{\sqrt{3}}, v$ variables take the value $\sqrt{3}, w(=0$ or 1$)$ variables take a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$ (if $w=1$, then

Since $(\sqrt{3} t+1) w=1043-4 v$ and $v$ is an integer, $w=1$, therefore, $-\frac{1}{\sqrt{3}}<t=\frac{1}{\sqrt{3}}(1042-4 v)<\sqrt{3}$, solving this gives $v=260, t=\frac{2}{\sqrt{3}}, u=1736$, thus the maximum value sought is
$$
\left(-\frac{1}{\sqrt{3}}\right)^{12} u+(\sqrt{3})^{12} v+t^{12}=\frac{1736+4096}{729}+729 \times 260=189548
$$