The straight line $y=ax+16$ intersects the graph of $y=x^3$ at $2$ distinct points. What is the value of $a$?
1. **Set up the equations:**
   The line \( y = ax + 16 \) intersects the curve \( y = x^3 \). To find the points of intersection, set the equations equal to each other:
   \[
   ax + 16 = x^3
   \]
   Rearrange this to form a polynomial equation:
   \[
   x^3 - ax - 16 = 0
   \]

2. **Determine the conditions for distinct points:**
   For the line to intersect the curve at 2 distinct points, the polynomial \( x^3 - ax - 16 = 0 \) must have exactly 2 distinct real roots. This implies that the polynomial has a double root and a single root.

3. **Find the derivative:**
   To find the double root, we need to find where the derivative of the polynomial is zero. The derivative of \( x^3 - ax - 16 \) is:
   \[
   \frac{d}{dx}(x^3 - ax - 16) = 3x^2 - a
   \]
   Set the derivative equal to zero to find the critical points:
   \[
   3x^2 - a = 0 \implies a = 3x^2
   \]

4. **Substitute back into the original equation:**
   Substitute \( a = 3x^2 \) into the original polynomial equation:
   \[
   x^3 - 3x^3 - 16 = 0 \implies -2x^3 - 16 = 0 \implies x^3 = -8 \implies x = -2
   \]

5. **Find the value of \( a \):**
   Substitute \( x = -2 \) back into \( a = 3x^2 \):
   \[
   a = 3(-2)^2 = 3 \cdot 4 = 12
   \]

6. **Verify the solution:**
   To ensure that the polynomial \( x^3 - 12x - 16 = 0 \) has exactly 2 distinct roots, we can check the discriminant or use the fact that the polynomial has a double root at \( x = -2 \) and another distinct root. The polynomial can be factored as:
   \[
   (x + 2)^2(x - 4) = 0
   \]
   This confirms that the polynomial has roots at \( x = -2 \) (double root) and \( x = 4 \) (single root).

The final answer is \( \boxed{12} \)