An awful identity We take place on $\mathbb C(x_1,...,x_r,x'_1,...,x'_p,u_0,...,u_r,u'_0,...,u'_p)$ with $r,p\in \mathbb N$

Show that :
  $$\displaystyle{\sum_{i=1}^r \left(   \frac{\prod_{j=0}^r (u_j-x_i) \prod_{j=1}^p (x'_j-x_i) }{  \prod_{1\leq j\leq r, j\neq i     }(x_j-x_i) \prod_{j=0}^p (u'_j-x_i)}      \right)}-\displaystyle{\sum_{i=0}^p \left(   \frac{\prod_{j=0}^r (u'_i-u_j) \prod_{j=1}^p (u'_i-x'_j) }{  \prod_{0\leq j\leq p, j\neq i     }(u'_i-u'_j) \prod_{j=1}^r (u'_i-x_j)}      \right)}=\displaystyle{\sum_{i=1}^p x'_i +\sum_{i=0}^r u_i -\sum_{i=1}^r x_i-\sum_{i=0}^p u'_i}$$

I have seen this exercise in an old book at the library. Unfortunately there was no indication how can I solve the problem. 
Anyway if someone can give me some ideas it will be greatful. 
NB: Sorry for the title, I have not had other ideas
 A: Idea:
Use the Lagrange interpolation formula for the $(r+p)$th degree polynomial
$$f(z)=\prod_{j=1}^r(z-x_j)\prod_{k=0}^p(z-u'_k)-\prod_{j=1}^p(z-x'_j)\prod_{k=0}^r(z-u_k),$$
interpolated at $r+p+1$ points $x_1,\ldots,x_r,u'_0,\ldots,u'_p$. Computing the coefficient of $z^{r+p}$ at both sides gives the identity you want to prove.

Proof:
Let us first recall the formula for the Lagrange interpolation polynomial
$$L(z)=\sum_{i=1}^{N} f_i\prod_{j\neq i}^{N}\frac{z-z_j}{z_i-z_j}.\tag{$\spadesuit$}$$
This is the unique $(N-1)$th degree polynomial passing through $N$ points
$(z_i,f_i)$ (with $i=1,\ldots, N$). 
Next let us consider a specific   polynomial (note that it also has degree $N-1$)
$$P(z)=\prod_{i=1}^{N}(z-z'_i)-\prod_{i=1}^{N}(z-z_i),\tag{$\clubsuit$}$$
which depends on $2N$ parameters $z_1,\ldots,z_{N}$, $z_1',\ldots,z_N'$. The values of this polynomial at $z=z_i$ are given by
$$P(z_i)=\prod_{j=1}^N(z_i-z'_j).$$
Hence, using ($\spadesuit$) with $f_i=P(z_i)$, we obtain the identity
$$\sum_{i=1}^N\prod_{k=1}^N(z_i-z'_k)\prod_{j\neq i}^{N}\frac{z-z_j}{z_i-z_j}=
\prod_{i=1}^{N}(z-z'_i)-\prod_{i=1}^{N}(z-z_i).\tag{$\diamondsuit$}$$
This is an equality between two polynomials. Now compute the coefficient in front of $z^{N-1}$ at both sides of ($\diamondsuit$). This leads to the identity
$$\sum_{i=1}^N\frac{\prod_{j=1}^N(z_i-z'_j)}{
\prod_{j\neq i}^N(z_i-z_j)}=
\sum_{i=1}^N z_i-\sum_{i=1}^Nz'_i.\tag{$\heartsuit$}$$
Finally, let us set $N=r+p+1$ and choose $\{z_i\}$, $\{z'_i\}$ in ($\heartsuit$) in the following way:
\begin{align*}
z_i&=\begin{cases}x_i & \text{for }i=1,\ldots,r,\\
u'_{i-r-1} & \text{for } i=r+1,\ldots,r+p+1,
\end{cases}\\
z'_i&=\begin{cases}u_i & \text{for }i=1,\ldots,r,\\
x'_{i-r-1} & \text{for } i=r+1,\ldots,r+p+1.
\end{cases}
\end{align*}
Then ($\heartsuit$) transforms into
\begin{align*}
\sum_{i=1}^r\frac{\prod_{j=1}^r(x_i-u_j)\prod_{j=0}^p
(x_i-x'_j)}{\prod_{j\neq i}^r(x_i-x_j)\prod_{j=0}^p(x_i-u'_j)}+
\sum_{i=0}^{p}\frac{
\prod_{j=1}^r(u'_i-u_j)\prod_{j=0}^p
(u'_i-x'_j)
}{\prod_{j=1}^r(u'_i-x_j)\prod_{j\neq i}^p(u'_i-u'_j)}=\\
=\sum_{i=1}^r x_i+\sum_{i=0}^p u'_i-\sum_{i=1}^r u_i-\sum_{i=0}^p x'_i,
\end{align*}
and this is nothing but the necessary identity. $\blacksquare$
