Birthday special 2

Using $\omega = e\displaystyle ^{\frac{2\pi i}{3}}$ and series expansion $e^x = \displaystyle \sum_{k = 0}^{\infty} \frac{x^k}{k!}$ we can combine these series via following relations

$A+B+C = e^x$ $A+\omega B + \omega^2 C = e^{\omega x}$ $A+\omega^2 B + \omega C = e^{\omega^2 x}$ Now $A^3+B^3+C^3-3ABC = (A+B+C)(A+\omega B + \omega^2 C)(A+\omega^2 B + \omega C)$ $= e^{(1+\omega+\omega^2)x} = e^{0} = 1$ i.e $A^3+B^3+C^3 = 1+3ABC$ Putting this value in the given relation we get $1+3ABC+8ABC = 2014$ Hence $ABC = \boxed{183}$