How can partial fractions be used for deductions? 
Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$
  . Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)}=a+b+c+d-p-q-r-s$ 

My Working
I was able to calculate partial fractions as follows,
$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)(x-a)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)(x-b)}+..$
But I cannot proceed to deduction part. Highly appreciated if someone can give me a hint to work this out. Thank you!
 A: You already got the correct partial fraction decomposition
$$
\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{1}{x-a} \\
+  \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{1}{x-b} \\
+  \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{1}{x-c} \\
+  \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{1}{x-d} \, .
$$
With the substitution $x = \frac 1y$ we get
$$ \tag{*}
\frac{(1-py)(1-qy)(1-ry)(1-sy)}{(1-ay)(1-by)(1-cy)(1-dy)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{y}{1-ay} \\
+  \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{y}{1-by} \\
+  \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{y}{1-cy} \\
+  \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{y}{1-dy} \, .
$$
For small $y$ the left-hand side has the development
$$
 \frac{1 - (p+q+r+s)y + O(y^2)}{1-(a+b+c+d)y + O(y^2)}
= (1 - (p+q+r+s)y + O(y^2))(1+(a+b+c+d)y + O(y^2) \\
= 1 + (a+b+c+d-p-q-r-s)y + O(y^2) 
$$
for $y \to 0$. On the right-hand side we have
$$
\frac{y}{1-ay} = y(1+ay + O(y^2)) = y + O(y^2) 
$$
and similarly for the other fractions $\frac{y}{1-by}$, $\frac{y}{1-cy}$
and $\frac{y}{1-dy}$.
Therefore a comparison of the $y$-terms (i.e. the derivates with respect
to $y$ at $y = 0$) in equation $(**)$ gives the desired identity
$$
a+b+c+d-p-q-r-s = \frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \\
+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \\
+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \, .
$$
A: This is $$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x-d}$$
The result should be this here
$$1+{\frac {{c}^{4}-{c}^{3}p-{c}^{3}q-{c}^{3}r-{c}^{3}s+{c}^{2}pq+{c}^{2
}pr+{c}^{2}ps+{c}^{2}qr+{c}^{2}qs+{c}^{2}rs-cpqr-cpqs-cprs-cqrs+pqrs}{
 \left( -c+a \right)  \left( -c+b \right)  \left( -d+c \right) 
 \left( x-c \right) }}+{\frac {-{d}^{4}+{d}^{3}p+{d}^{3}q+{d}^{3}r+{d}
^{3}s-{d}^{2}pq-{d}^{2}pr-{d}^{2}ps-{d}^{2}qr-{d}^{2}qs-{d}^{2}rs+dpqr
+dpqs+dprs+dqrs-pqrs}{ \left( x-d \right)  \left( -d+a \right) 
 \left( -d+b \right)  \left( -d+c \right) }}+{\frac {-{b}^{4}+{b}^{3}p
+{b}^{3}q+{b}^{3}r+{b}^{3}s-{b}^{2}pq-{b}^{2}pr-{b}^{2}ps-{b}^{2}qr-{b
}^{2}qs-{b}^{2}rs+bpqr+bpqs+bprs+bqrs-pqrs}{ \left( x-b \right) 
 \left( -b+a \right)  \left( -c+b \right)  \left( -d+b \right) }}+{
\frac {{a}^{4}-{a}^{3}p-{a}^{3}q-{a}^{3}r-{a}^{3}s+{a}^{2}pq+{a}^{2}pr
+{a}^{2}ps+{a}^{2}qr+{a}^{2}qs+{a}^{2}rs-apqr-apqs-aprs-aqrs+pqrs}{
 \left( -b+a \right)  \left( -c+a \right)  \left( -d+a \right) 
 \left( x-a \right) }}
$$
