We solve this problem by reducing it to something more manageable.
Let x=2a-b, y= 2b-c, z=2c-a.
Then x+y+z= 2a-b+2b-c+2c-a = a+b+c = 0, by assumption.
So we have to show that if x+y+z=0, then x^3+y^3+z^3-3xyz=0.
But
(x+y+z)^3 = x^3+y^3+z^3 +3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z) +6xyz...