([b]3[/b]) Let $ \ell$ be the line through $ (0,0)$ and tangent to the curve $ y \equal{} x^3 \plus{} x \plus{} 16$. Find the slope of $ \ell$.
1. **Identify the curve and its derivative:**
   The given curve is \( y = x^3 + x + 16 \). To find the slope of the tangent line at any point on this curve, we need to compute the derivative of \( y \) with respect to \( x \):
   \[
   \frac{dy}{dx} = \frac{d}{dx}(x^3 + x + 16) = 3x^2 + 1
   \]

2. **Set up the equation for the tangent line:**
   The line \( \ell \) passes through the origin \((0,0)\) and is tangent to the curve at some point \((a, a^3 + a + 16)\). The slope of the tangent line at this point is given by the derivative at \( x = a \), which is \( 3a^2 + 1 \).

3. **Equation of the line through the origin:**
   The equation of the line through the origin with slope \( m \) is \( y = mx \). Since the line is tangent to the curve at \( x = a \), the slope \( m \) must be equal to the derivative at \( x = a \):
   \[
   m = 3a^2 + 1
   \]

4. **Find the point of tangency:**
   The point of tangency \((a, a^3 + a + 16)\) must lie on the line \( y = mx \). Therefore, substituting \( y = a^3 + a + 16 \) and \( y = mx \) into the equation of the line, we get:
   \[
   a^3 + a + 16 = (3a^2 + 1)a
   \]
   Simplifying this equation:
   \[
   a^3 + a + 16 = 3a^3 + a
   \]
   \[
   16 = 2a^3
   \]
   \[
   a^3 = 8
   \]
   \[
   a = 2
   \]

5. **Calculate the slope at the point of tangency:**
   The slope of the tangent line at \( x = 2 \) is:
   \[
   m = 3(2)^2 + 1 = 3 \cdot 4 + 1 = 12 + 1 = 13
   \]

Conclusion:
The slope of the line \( \ell \) is \( 13 \).

The final answer is \(\boxed{13}\).