Let $a$ and $b$ satisfy the conditions $\begin{cases}
a^3 - 6a^2 + 15a = 9 \\
b^3 - 3b^2 + 6b = -1 \end{cases}$ .
The value of $(a - b)^{2014}$ is:

(A): $1$, (B): $2$, (C): $3$, (D): $4$, (E) None of the above.
1. We start with the given system of equations:
   \[
   \begin{cases}
   a^3 - 6a^2 + 15a = 9 \\
   b^3 - 3b^2 + 6b = -1 
   \end{cases}
   \]

2. Rewrite the first equation:
   \[
   a^3 - 6a^2 + 15a - 9 = 0
   \]
   We will try to express this in a form that might reveal a relationship with \(b\). Consider the transformation \(a = x + 1\):
   \[
   (x+1)^3 - 6(x+1)^2 + 15(x+1) - 9 = 0
   \]
   Expanding each term:
   \[
   (x+1)^3 = x^3 + 3x^2 + 3x + 1
   \]
   \[
   6(x+1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6
   \]
   \[
   15(x+1) = 15x + 15
   \]
   Substituting these into the equation:
   \[
   x^3 + 3x^2 + 3x + 1 - 6x^2 - 12x - 6 + 15x + 15 - 9 = 0
   \]
   Simplifying:
   \[
   x^3 + 3x^2 - 6x^2 + 3x - 12x + 15x + 1 - 6 + 15 - 9 = 0
   \]
   \[
   x^3 - 3x^2 + 6x + 1 = 0
   \]
   This simplifies to:
   \[
   x^3 - 3x^2 + 6x + 1 = 0
   \]
   Notice that this is exactly the second equation with \(x = b\):
   \[
   b^3 - 3b^2 + 6b + 1 = 0
   \]
   Therefore, we have \(x = b\) and \(a = b + 1\).

3. Now, we need to find the value of \((a - b)^{2014}\):
   \[
   a - b = (b + 1) - b = 1
   \]
   Therefore:
   \[
   (a - b)^{2014} = 1^{2014} = 1
   \]

The final answer is \(\boxed{1}\)