Identity with roots of $\frac{x}{x-p}+\frac{x}{x-q}+\frac{x}{x-r}+(x-s)=0$. If the roots of $\frac{x}{x-p}+\frac{x}{x-q}+\frac{x}{x-r}+(x-s)=0$ are $\alpha, \beta, \gamma, \delta$, I want to show the identity:
$$ \frac{p^2}{(p-\alpha)(p-\beta)(p-\gamma)(p-\delta)}+\frac{q^2}{(q-\alpha)(q-\beta)(q-\gamma)(q-\delta)}+\frac{r^2}{(r-\alpha)(r-\beta)(r-\gamma)(r-\delta)}=0.$$
I am profoundly stuck on how to get the denominators to multiply and eliminate $s$. By the nature of $\alpha$ being a root I know that $\frac{\alpha}{\alpha-p}+\frac{\alpha}{\alpha-q}+\frac{\alpha}{\alpha-r}+(\alpha-s)=0$. I then wipe the denominators to get $\alpha(\alpha-q)(\alpha-r)+\alpha(\alpha-p)(\alpha-r)+\alpha(\alpha-p)(\alpha-q)+(\alpha-s)(\alpha-p)(\alpha-q)(\alpha-r)=0$. However this doesn't help at all because I am getting a product rotating between $p,q,r,s$ instead of $\alpha, \beta, \gamma, \delta$.
Could someone give me some leads on how to complete this question?
 A: $$\frac{x}{x-p}+\frac{x}{x-q}+\frac{x}{x-r}+(x-s)=0$$
Making common denominator gives
$$\frac{x(x-q)(x-r)+x(x-p)(x-r)+x(x-p)(x-q)+(x-p)(x-q)(x-r)(x-s)}{(x-p)(x-q)(x-r)}=0$$
Let numerator of LHS is $f(x)$ and denominator is $g(x)$.
All roots of $\frac{f(x)}{g(x)}=0$ are roots of $f(x)=0$, then $f(x)=0$ has at least four roots $\alpha$, $\beta$, $\gamma$, $\delta$.
Note that $$f(x)=x(x-q)(x-r)+x(x-p)(x-r)+x(x-p)(x-q)+(x-p)(x-q)(x-r)(x-s)$$ is polynomial of 4th degree with senior coefficient $a_4=1$. Then it has at most 4 roots, and these roots are $\alpha$, $\beta$, $\gamma$, $\delta$. Then $$f(x)=a_4(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$$
Then for any $x$ $$x(x-q)(x-r)+x(x-p)(x-r)+x(x-p)(x-q)+(x-p)(x-q)(x-r)(x-s)=\\=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$$
Using this equality for $x=p$, $x=q$, $x=r$ gives
$$p(p-q)(p-r)=(p-\alpha)(p-\beta)(p-\gamma)(p-\delta)$$
$$q(q-p)(q-r)=(q-\alpha)(q-\beta)(q-\gamma)(q-\delta)$$
$$r(r-p)(r-q)=(r-\alpha)(r-\beta)(r-\gamma)(r-\delta)$$
Putting LHS of these equalities into identity which is necessary to prove instead of RHS expressions gives
$$\frac{p^2}{p(p-q)(p-r)}+\frac{q^2}{q(q-p)(q-r)}+\frac{r^2}{r(r-p)(r-q)}=0$$
Reducing and making common denominator gives
$$\frac{p(q-r)}{(p-q)(p-r)(q-r)}+\frac{-q(p-r)}{(p-q)(p-r)(q-r)}+\frac{r(p-q)}{(p-q)(p-r)(q-r)}=0$$
$$p(q-r)-q(p-r)+r(p-q)=0$$
$$pq-pr-pq-qr+pr-qr=0$$
$$0=0$$
