Show that a given polynomial can't have a multiple root occurring more $n-1$ times Question:

Let  $x_{1},x_{2},\dots,x_{n}$ be a complex numbers such $x_{i}\neq x_{j},\forall i\neq j$. Show that the following polynomial
  $$p(x)=(x-x_{1})^2(x-x_{2})^2\cdots (x-x_{n})^2\cdot\left(\dfrac{1}{(x-x_{1})^2}+\dfrac{1}{(x-x_{2})^2}+\cdots+\dfrac{1}{(x-x_{n})^2}\right)$$
  can't have a root $b$ which is a multiple root occurring more $n-1$ times.

My idea: when $n=2$, then
$$p(x)=(x-x_{1})^2(x-x_{2})^2\left(\dfrac{1}{(x-x_{1})^2}+\dfrac{1}{(x-x_{2})^2}\right)=(x-x_{1})^2+(x-x_{2})^2$$
since $x_{1},x_{2}$ is different, so
$p(x)=0$ can't have a root which a multiple root occurring more $1$ times.
But for other cases I can't. Thank you.
 A: We prove by contradicts:
If $p(x)$ have a root $b$ which multiple root occurring more $n-1$ times, then :
$$p(b)=p'(b)=\cdots=p^{(n-1)}(b)=0$$
$$b\neq x_1,x_2,\cdots,x_n$$ 
Denote $q(x)=(x-x_1)^2(x-x_2)^2\cdots(x-x_n)^2$, then as $q(x)\neq0$ in some $\epsilon$-neighborhood of $b$, we consider $\frac{p(x)}{q(x)}=\sum_{i=1}^n\frac{1}{(x-x_i)^2}$:
$$\frac{p(x)}{q(x)}=(\frac{p(x)}{q(x)})'=\cdots(\frac{p(x)}{q(x)})^{(n-1)}=0\,\text{at}\,x=b$$
This imply that
$$\sum_{i=1}^n\frac{1}{(b-x_i)^2}=\sum_{i=1}^n\frac{1}{(b-x_i)^3}=\sum_{i=1}^n\frac{1}{(b-x_i)^{n+1}}=0$$
where $$\frac{1}{b-x_1}\neq\frac{1}{b-x_2}\neq\cdots\neq\frac{1}{b-x_n}\neq0$$
Consider the Vandermonde matrix:
$$V=\begin{bmatrix}
\frac{1}{(b-x_1)^2} & \frac{1}{(b-x_2)^2} & \dots & \frac{1}{(b-x_n)^2}\\
\frac{1}{(b-x_1)^3} & \frac{1}{(b-x_2)^3} & \dots & \frac{1}{(b-x_n)^3}\\
\vdots & \vdots &  \ddots &\vdots \\
\frac{1}{(b-x_1)^{n+1}} & \frac{1}{(b-x_2)^{n+1}} &  \dots & \frac{1}{(b-x_n)^{n+1}}
\end{bmatrix}$$
Then $\det V\neq0$ and on the other hand, there exist a vector $v=[1,1,\cdots,1]^t$ such that $Vv=0$, which makes the contradicts.
