TITLE: The coefficient of a specific monomial of the following polynomial
QUESTION [7 upvotes]: Let the real polynomial 
$$f_{a,b,c}(x_1,x_2,x_3)=(x_1-x_2)^{2a+1}(x_2-x_3)^{2b+1}(x_3-x_1)^{2c+1},$$
where $a,b,c$ are nonnegative integers.
Let $m_{a,b,c}$ be the coefficient of the monomial $x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$ in the expansion of $f_{a,b,c}(x_1,x_2,x_3)$.
It is easy to see $m_{a,b,c}=0$ when two of $a,b,c$ are equal. I want to ask whether $m_{a,b,c}=0$ or not when $a,b,c$ are pairwise unequal.

REPLY [17 votes]: This coefficient $L$ is a constant term of the Laurent polynomial $g(x_1,x_2,x_3)=f(x_1,x_2,x_3)/x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$, this guy $g$ satisfies $g(x_1,x_2,x_3)=-g(1/x_1,1/x_2,1/x_3)$, thus $L=-L$.

REPLY [4 votes]: This answer is really just a convoluted edition of Fedor's answer.  By Cauchy's theorem,
$$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi
   \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(a+c)\theta_1+i(a+b)\theta_2+i(b+c)\theta_3}}
\,d\theta_1\,d\theta_2\,d\theta_3.$$
Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.
To obtain the above integral start with the contour integral
$$\frac{1}{(2\pi i)^3} \oint\oint\oint \frac{(x_1-x_2)^{2a+1} (x_2-x_3)^{2b+1} (x_3-x_1)^{2c+1}}{x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}} \,dx_1dx_2dx_3 $$
that is just Cauchy's formula applied three times. Now take the contours to be unit circles $x_j=e^{i\theta_j}$ for $j=1,2,3$.