已知 \(x_1, x_2, \dots, x_9\) 满足方程组
\[
\begin{cases}
\dfrac{x_1}{11} + \dfrac{x_2}{12} + \dots + \dfrac{x_9}{19} = 1 \\
\dfrac{x_1}{21} + \dfrac{x_2}{22} + \dots + \dfrac{x_9}{29} = 1 \\
\dfrac{x_1}{31} + \dfrac{x_2}{32} + \dots + \dfrac{x_9}{39} = 1 \\
\dfrac{x_1}{41} + \dfrac{x_2}{42} + \dots + \dfrac{x_9}{49} = 1 \\
\dfrac{x_1}{51} + \dfrac{x_2}{52} + \dots + \dfrac{x_9}{59} = 1 \\
\dfrac{x_1}{61} + \dfrac{x_2}{62} + \dots + \dfrac{x_9}{69} = 1 \\
\dfrac{x_1}{71} + \dfrac{x_2}{72} + \dots + \dfrac{x_9}{79} = 1 \\
\dfrac{x_1}{81} + \dfrac{x_2}{82} + \dots + \dfrac{x_9}{89} = 1 \\
\dfrac{x_1}{91} + \dfrac{x_2}{92} + \dots + \dfrac{x_9}{99} = 1 \\
\end{cases}
\]
试求 \(x_1 + x_2 + \dots + x_9\).

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试求所有的实系数多项式 \(p(x)\), 使得对满足 \(ab + bc + ca = 0\) 的所有实数 \(a、b、c\), 均有:
\[
p(a - b) + p(b - c) + p(c - a) = 2p(a + b + c).
\]

已知数列 \(\{a_n\}\) 和 \(\{b_n\}\), 其中 \(n \ge 0\) (即两个数列的首项分别为 \(a_0\) 和 \(b_0\)). 若满足
\[
b_n = \sum_{k=0}^n (-1)^k C_n^k a_k
\,, \quad \forall n \in \mathbb N.
\]
求证:
\[
a_n = \sum_{k=0}^n (-1)^k C_n^k b_k
\,, \quad \forall n \in \mathbb N.
\]

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