The graph of the function $f(x)=x^n+a_{n-1}x_{n-1}+\ldots +a_1x+a_0$ (where $n>1$) intersects the line $y=b$ at the points $B_1,B_2,\ldots ,B_n$ (from left to right), and the line $y=c\ (c\not= b)$ at the points $C_1,C_2,\ldots ,C_n$ (from left to right). Let $P$ be a point on the line $y=c$, to the right to the point $C_n$. Find the sum
\[\cot (\angle B_1C_1P)+\ldots +\cot (\angle B_nC_nP) \]
1. Consider the polynomial function \( f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \) where \( n > 1 \). The graph of this function intersects the line \( y = b \) at points \( B_1, B_2, \ldots, B_n \) and the line \( y = c \) at points \( C_1, C_2, \ldots, C_n \).

2. Let \( b_1, b_2, \ldots, b_n \) be the roots of the equation \( f(x) = b \) and \( c_1, c_2, \ldots, c_n \) be the roots of the equation \( f(x) = c \). These roots represent the x-coordinates of the points \( B_i \) and \( C_i \) respectively.

3. We need to find the sum \( \cot (\angle B_1C_1P) + \ldots + \cot (\angle B_nC_nP) \) where \( P \) is a point on the line \( y = c \) to the right of \( C_n \).

4. Without loss of generality, assume \( c > b \). The cotangent of the angle \( \angle B_iC_iP \) can be expressed as:
   \[
   \cot (\angle B_iC_iP) = \frac{B_iC_i}{C_iP}
   \]
   where \( B_iC_i = b_i - c_i \) and \( C_iP \) is the horizontal distance from \( C_i \) to \( P \).

5. Since \( P \) is to the right of \( C_n \), the horizontal distance \( C_iP \) is the same for all \( i \) and can be denoted as \( d \). Thus,
   \[
   \cot (\angle B_iC_iP) = \frac{b_i - c_i}{d}
   \]

6. Summing these cotangents, we get:
   \[
   \sum_{i=1}^n \cot (\angle B_iC_iP) = \sum_{i=1}^n \frac{b_i - c_i}{d} = \frac{1}{d} \sum_{i=1}^n (b_i - c_i)
   \]

7. The sum of the roots of the polynomial \( f(x) = k \) (where \( k \) is any constant) is given by Vieta's formulas as \( -\frac{a_{n-1}}{1} = -a_{n-1} \). Therefore,
   \[
   \sum_{i=1}^n b_i = -a_{n-1} \quad \text{and} \quad \sum_{i=1}^n c_i = -a_{n-1}
   \]

8. Substituting these sums into the equation, we get:
   \[
   \sum_{i=1}^n (b_i - c_i) = (-a_{n-1}) - (-a_{n-1}) = 0
   \]

9. Therefore,
   \[
   \sum_{i=1}^n \cot (\angle B_iC_iP) = \frac{1}{d} \cdot 0 = 0
   \]

10. The same result holds if \( c < b \) by a similar argument.

Thus, we conclude that:
\[
\cot (\angle B_1C_1P) + \ldots + \cot (\angle B_nC_nP) = 0
\]

The final answer is \( \boxed{ 0 } \)