We are given that $n>3$ and we have to find the sum of the series given by: $$S=xyz\binom{n}{0}-(x-1)(y-1)(z-1)\binom{n}{1}+...+(-1)^n(x-n)(y-n)(z-n)\binom{n}{n}$$
I figured out that the general term is $$t(r)=(-1)^r(x-r)(y-r)(z-r)\binom{n}{r}$$ but I see no obvious manipulations between the terms nor does any particular series strike my mind.
Can someone provide an approach? Any help would be appreciated.
The given sum is zero for $n>3$. Note that $$(-1)^r(x-r)(y-r)(z-r)\binom{n}{r}$$ is a linear combination of $\binom{n}{r}r^k(-1)^r$ with $k=0,1,2,3$. Moreover, for $0\leq k<n$ we have that $$\sum_{r=0}^n\binom{n}{r}r^k(-1)^r=0$$ (for a detailed proof see Simplifying $\sum_{r = 0}^{n} {{n}\choose{r}}r^k(-1)^r$ ).