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

[i]Proposed by Evan Chen[/i]
1. **Define the series and their sums:**
   \[
   A = \sum_{k=0}^{\infty} \frac{x^{3k}}{(3k)!}, \quad
   B = \sum_{k=0}^{\infty} \frac{x^{3k+1}}{(3k+1)!}, \quad
   C = \sum_{k=0}^{\infty} \frac{x^{3k+2}}{(3k+2)!}.
   \]

2. **Key Claim:**
   We claim that:
   \[
   A^3 + B^3 + C^3 - 3ABC = 1
   \]
   for all \( x \).

3. **Differentiate the expression:**
   Let \( A(x), B(x), C(x) \) denote the given sums. We need to show that the derivative of \( A^3 + B^3 + C^3 - 3ABC \) is zero. This implies that the expression is constant.

   Compute the derivatives:
   \[
   \frac{d}{dx} A(x) = \sum_{k=1}^{\infty} \frac{3k x^{3k-1}}{(3k)!} = \sum_{k=1}^{\infty} \frac{x^{3k-1}}{(3k-1)!} = B(x)
   \]
   \[
   \frac{d}{dx} B(x) = \sum_{k=0}^{\infty} \frac{(3k+1) x^{3k}}{(3k+1)!} = \sum_{k=0}^{\infty} \frac{x^{3k}}{(3k)!} = A(x)
   \]
   \[
   \frac{d}{dx} C(x) = \sum_{k=0}^{\infty} \frac{(3k+2) x^{3k+1}}{(3k+2)!} = \sum_{k=0}^{\infty} \frac{x^{3k+1}}{(3k+1)!} = B(x)
   \]

4. **Apply the chain rule:**
   \[
   \frac{d}{dx} (A^3 + B^3 + C^3 - 3ABC) = 3A^2 \frac{dA}{dx} + 3B^2 \frac{dB}{dx} + 3C^2 \frac{dC}{dx} - 3 \left( \frac{dA}{dx}BC + A\frac{dB}{dx}C + AB\frac{dC}{dx} \right)
   \]
   Substitute the derivatives:
   \[
   = 3A^2 B + 3B^2 A + 3C^2 B - 3 (B B C + A A C + A B B)
   \]
   Simplify:
   \[
   = 3A^2 B + 3B^2 A + 3C^2 B - 3 (B^2 C + A^2 C + A B^2)
   \]
   Notice that each term cancels out:
   \[
   = 0
   \]

5. **Initial condition:**
   Evaluate at \( x = 0 \):
   \[
   A(0) = 1, \quad B(0) = 0, \quad C(0) = 0
   \]
   \[
   A(0)^3 + B(0)^3 + C(0)^3 - 3A(0)B(0)C(0) = 1^3 + 0^3 + 0^3 - 3 \cdot 1 \cdot 0 \cdot 0 = 1
   \]

6. **Conclusion:**
   Since the derivative is zero and the initial condition is 1, we have:
   \[
   A^3 + B^3 + C^3 - 3ABC = 1
   \]

7. **Given equation:**
   \[
   A^3 + B^3 + C^3 + 8ABC = 2014
   \]
   Substitute the key claim:
   \[
   1 + 11ABC = 2014
   \]
   Solve for \( ABC \):
   \[
   11ABC = 2013
   \]
   \[
   ABC = \frac{2013}{11} = 183
   \]

The final answer is \( \boxed{183} \).