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This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ 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 $j=1,2,3$.

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 $j=1,2,3$.

This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.

This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ 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 $j=1,2,3$.

This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1-i\theta_2})^{2a+1}(e^{i\theta_2-i\theta_3})^{2b+1}(e^{i\theta_3-i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.

This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(2a+2b+2c)}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.