Let $ f(x) = x^3 + ax + b $, with $ a \ne b $, and suppose the tangent lines to the graph of $f$ at $x=a$ and $x=b$ are parallel. Find $f(1)$.
1. Given the function \( f(x) = x^3 + ax + b \), we need to find \( f(1) \) under the condition that the tangent lines to the graph of \( f \) at \( x = a \) and \( x = b \) are parallel.
2. The condition for the tangent lines to be parallel is that their slopes must be equal. The slope of the tangent line to the graph of \( f \) at any point \( x \) is given by the derivative \( f'(x) \).
3. Compute the derivative of \( f(x) \):
   \[
   f'(x) = \frac{d}{dx}(x^3 + ax + b) = 3x^2 + a
   \]
4. Set the slopes at \( x = a \) and \( x = b \) equal to each other:
   \[
   f'(a) = f'(b)
   \]
   Substituting the derivative, we get:
   \[
   3a^2 + a = 3b^2 + a
   \]
5. Simplify the equation:
   \[
   3a^2 + a - 3b^2 - a = 0
   \]
   \[
   3a^2 = 3b^2
   \]
   \[
   a^2 = b^2
   \]
6. Since \( a \neq b \), the only solution is \( a = -b \).
7. Now, we need to find \( f(1) \):
   \[
   f(1) = 1^3 + a \cdot 1 + b = 1 + a + b
   \]
8. Substitute \( b = -a \) into the expression:
   \[
   f(1) = 1 + a - a = 1
   \]

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