For natural numbers $a_1,\dots,a_n$, Freeman Dyson conjectured (and it was eventually proven) that the Laurent polynomial $ \prod_{i,j=1\atop i\neq j}^n\left(1-\frac{x_i}{x_j}\right)^{a_i} $ has constant term the multinomial coefficient $\binom{a_1+\cdots+a_n}{a_1,\dots,a_n}$.
A.C. Dixon proved Dixon's Identity: $ \sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}=\frac{(a+b+c)!}{a!b!c!}. $
This last quantity is just the multinomial coefficient $\binom{a+b+c}{a,b,c}$. So letting $a_1=a,a_2=b,a_3=c$, it should be the case by Dyson's conjecture that $ \prod_{i,j=1\atop i\neq j}^3\left(1-\frac{x_i}{x_j}\right)^{a_i} $ that is, $ \left(1-\frac{x_1}{x_2}\right)^a\left(1-\frac{x_1}{x_3}\right)^a\left(1-\frac{x_2}{x_1}\right)^b\left(1-\frac{x_2}{x_3}\right)^b\left(1-\frac{x_3}{x_1}\right)^c\left(1-\frac{x_3}{x_2}\right)^c $ has constant term $\binom{a+b+c}{a,b,c}$. Does anyone see a clever way to conclude that the constant term is also calculated as $ \sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}$ to get another proof of the result? Many thanks.
My guess is that one will want to "choose" say the same number of factors say $-\frac{x_1}{x_2}$ from $(1-\frac{x_1}{x_2})^a$ as the number of factors $-\frac{x_2}{x_1}$ from $(1-\frac{x_2}{x_1})^b$, when expanding. So I assume the sum fro $-a$ to $a$ will count how many ways there are to choose a carefully from the six terms in the product to get a constant when it's all multiplied out.