z(a, a) = (z(a)^2 - z(2a)) / 2
z(a)^3 = \sum_{x, y, z} 1/x^ay^az^a = 6 z(a,a,a) (x, y, z distinct) + 3 z(2a,a) + 3 z(a,2a) (two of x, y, z equal) + z(3a) (all equal)
z(2a,a) + z(a, 2a) = \sum_{x,y, x!=y} 1/x^(2a)y^a = z(2a)z(a) - z(3a)
Therefore, z(a)^3 = 6z(a,a,a) + 3 z(2a)z(a) - 2 z(3a)