The equation $ ax^3\plus{}bx^2\plus{}cx\plus{}d\equal{}0$ has three distinct solutions. How many distinct solutions does the following equation have:

$ 4(ax^3\plus{}bx^2\plus{}cx\plus{}d)(3ax\plus{}b)\equal{}(3ax^2\plus{}2bx\plus{}c)^2?$
1. Let \( f(x) = ax^3 + bx^2 + cx + d \). Given that \( f(x) = 0 \) has three distinct solutions, denote these solutions by \( \alpha, \beta, \gamma \) where \( \alpha < \beta < \gamma \).

2. The derivative of \( f(x) \) is:
   \[
   f'(x) = 3ax^2 + 2bx + c
   \]

3. The second derivative of \( f(x) \) is:
   \[
   f''(x) = 6ax + 2b
   \]

4. The third derivative of \( f(x) \) is:
   \[
   f'''(x) = 6a
   \]

5. The given equation is:
   \[
   4(ax^3 + bx^2 + cx + d)(3ax + b) = (3ax^2 + 2bx + c)^2
   \]

6. Substitute \( f(x) \) and its derivatives into the equation:
   \[
   4f(x)f'(x) = (f'(x))^2
   \]

7. Simplify the equation:
   \[
   4f(x)f'(x) = (3ax^2 + 2bx + c)^2
   \]

8. Rearrange the equation:
   \[
   2f(x)f''(x) = (f'(x))^2
   \]

9. Define a new function \( g(x) \):
   \[
   g(x) = 2f(x)f''(x) - (f'(x))^2
   \]

10. Differentiate \( g(x) \):
    \[
    g'(x) = 2f(x)f'''(x) = 12a^2(x - \alpha)(x - \beta)(x - \gamma)
    \]

11. Evaluate \( g(x) \) at the roots \( \alpha, \beta, \gamma \):
    \[
    g(\alpha) = 2f(\alpha)f''(\alpha) - (f'(\alpha))^2 = -(f'(\alpha))^2 < 0
    \]
    \[
    g(\beta) = 2f(\beta)f''(\beta) - (f'(\beta))^2 = -(f'(\beta))^2 < 0
    \]
    \[
    g(\gamma) = 2f(\gamma)f''(\gamma) - (f'(\gamma))^2 = -(f'(\gamma))^2 < 0
    \]

12. Since \( g(x) \) changes sign at each of the roots \( \alpha, \beta, \gamma \), the equation \( g(x) = 0 \) has exactly two distinct solutions.

The final answer is \( \boxed{2} \).