Let \(x\) be a positive real number. Define
\[
A =\displaystyle\sum_{k=0}^{\infty} \dfrac{x^{3k}}{(3k)!}, \quad
B = \displaystyle\sum_{k=0}^{\infty} \dfrac{x^{3k+1}}{(3k+1)!}, \quad\text{and}\quad
C = \displaystyle\sum_{k=0}^{\infty} \dfrac{x^{3k+2}}{(3k+2)!}.
\] Given that \(A^3+B^3+C^3 + 8ABC = 2014\) , compute \(ABC\) .

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