Real numbers only

We can expand and separate the expression into:

$\displaystyle \sum_{cyc}^{a, b, c} a(b-c)^3=\displaystyle \sum_{cyc}^{a, b, c} ab^3-3ab^2 c+3abc^2-ac^3$

$\displaystyle \sum_{cyc}^{a, b, c} a(b-c)^3=\displaystyle \sum_{cyc}^{a, b, c} (ab^3-ac^3) - 3abc \displaystyle \sum_{cyc}^{a, b, c} (c-b)$

$\displaystyle \sum_{cyc}^{a, b, c} a(b-c)^3=ab^3-ba^3+bc^3-cb^3+ca^3-ac^3 - (3abc)(0) =\displaystyle \sum_{cyc}^{a, b, c} ab(b-a)(b+a)$

$\displaystyle \sum_{cyc}^{a, b, c} a(b-c)^3=\displaystyle \sum_{cyc}^{a, b, c} ab(b-a)(-c)=-abc \displaystyle \sum_{cyc}^{a, b, c} (b-a)$

$\displaystyle \sum_{cyc}^{a, b, c} a(b-c)^3=(-abc)(0) = \boxed{0}$