# A frightening sum [duplicate]

Let $x_1,\ldots,x_r,y_1,\ldots,y_p,z_0,\ldots,z_r,t_0,\ldots,t_p$ be complex numbers. Let $A$ be the ring generated by these numbers.

Prove the following holds in $\mathbb C(A)$.

$$\begin{multline*}\sum_{i=1}^r \left(\prod_{j=0}^r(z_j-x_i)\prod_{1\leq j\leq r; j\neq i}\left({x_j-x_i}\right)^{-1}\prod_{j=1}^r(y_j-x_i)\prod_{j=0}^p\left({t_j-x_i} \right)^{-1}\right)\\-\sum_{i=0}^p \left(\prod_{j=0}^r(t_i-z_j)\prod_{j=1}^r {(t_i-x_j)}^{-1}\prod_{j=1}^p(t_i-y_j)\prod_{0\leq j \leq p; j\neq i}\left({t_i-t_j}\right)^{-1}\right) \\=-\left(\sum_{i=1}^rx_i-\sum_{i=0}^rz_i \right)+\left(\sum_{i=1}^py_i-\sum_{i=0}^pt_i \right)\end{multline*}$$

This was asked at an oral examination. I post it out of curiosity (no idea how it should be approached).

EDIT It is a duplicate of An awful identity

• I'm curious how to interpret the products such as $\prod_{j=0}^r (t_i-z_j)$ if $r > p$..?
– Tom
Jun 10 '14 at 12:28
• This was asked in an oral examination?? Where? In the "Inquisition-Hell University, Dept. of Mental Torture and Pointless Cruelty"?? More than a disturbance I sense a hurricane in the force... Jun 10 '14 at 12:30
• @Tom It makes perfect sense though. $i$ is fixed from the sum outside and only $j$ varies inside the product. Jun 10 '14 at 12:34
• @DonAntonio This is the place ens.fr/a-propos/l-ecole Jun 10 '14 at 12:37
• I seem @G.T.R: a top university's mathematics dept. This question belongs perhaps to the realm of applied mathematics. Jun 10 '14 at 12:47

I don't think this is true, if you would set $x_i=z_i=t_i$, the left hand side would be $0$ (because at least one factor in the product would be $0$) and the right would not...