Let $\alpha < 0 < \beta$ and consider the polynomial $f(x) = x(x-\alpha)(x-\beta)$. Let $S$ be the set of real numbers $s$ such that $f(x) - s$ has three different real roots. For $s\in S$, let $p(x)$ the product of the smallest and largest root of $f(x)-s$. Determine the smallest possible value that $p(s)$ for  $s\in S$.
1. **Identify the roots of \( f(x) = x(x - \alpha)(x - \beta) \):**
   The roots of \( f(x) \) are \( 0 \), \( \alpha \), and \( \beta \).

2. **Condition for \( f(x) - s \) to have three different real roots:**
   For \( f(x) - s \) to have three different real roots, \( s \) must be such that the polynomial \( f(x) - s = x(x - \alpha)(x - \beta) - s \) intersects the x-axis at three distinct points.

3. **Case \( s = 0 \):**
   If \( s = 0 \), the roots are \( 0 \), \( \alpha \), and \( \beta \). The product of the smallest and largest root is \( \alpha \beta \).

4. **Case \( s \neq 0 \):**
   If \( s \neq 0 \), then \( 0 \) is not a root of \( f(x) - s \). Let \( p \leq r \leq q \) be the roots of \( f(x) - s = 0 \). Then, \( p(s) = pq = \frac{s}{r} \).

5. **Expression for \( \frac{s}{r} \):**
   Since \( f(r) - s = 0 \), we have:
   \[
   \frac{s}{r} = \frac{f(r)}{r} = r^2 - (\alpha + \beta)r + \alpha \beta
   \]

6. **Minimizing \( \frac{s}{r} \):**
   We need to find:
   \[
   \min \left\{ \alpha \beta, \min_{r \in R} \left\{ r^2 - (\alpha + \beta)r + \alpha \beta \right\} \right\}
   \]
   where \( R \) is the set of admissible \( r \).

7. **Finding the minimum of \( r^2 - (\alpha + \beta)r + \alpha \beta \):**
   The quadratic function \( r^2 - (\alpha + \beta)r + \alpha \beta \) achieves its minimum at:
   \[
   r = \frac{\alpha + \beta}{2}
   \]

8. **Checking if \( r = \frac{\alpha + \beta}{2} \) is admissible:**
   We need to show that \( r = \frac{\alpha + \beta}{2} \) induces a real \( s \) such that \( f(x) - s \) has three real roots and \( \frac{\alpha + \beta}{2} \) is the central root.

9. **Using Vieta's formulas:**
   By Vieta's formulas, the roots \( p \), \( q \), and \( \frac{\alpha + \beta}{2} \) satisfy:
   \[
   p + q + \frac{\alpha + \beta}{2} = \alpha + \beta
   \]
   and
   \[
   pq \left( \frac{\alpha + \beta}{2} \right) = s
   \]

10. **Solving for \( p \) and \( q \):**
    The quadratic equation for \( p \) and \( q \) is:
    \[
    t^2 - \left( \frac{\beta + \alpha}{2} \right)t - \left( \frac{\beta - \alpha}{2} \right)^2 = 0
    \]
    Solving this, we get:
    \[
    t = \frac{\beta + \alpha}{4} \pm \frac{1}{2} \sqrt{\left( \frac{\beta + \alpha}{2} \right)^2 + 4 \left( \frac{\beta - \alpha}{2} \right)^2}
    \]
    which are real values.

11. **Conclusion:**
    Since \( r = \frac{\alpha + \beta}{2} \) is admissible, we have:
    \[
    \min_{s \in S} p(s) = \min \left\{ \alpha \beta, -\frac{1}{4} (\beta - \alpha)^2 \right\}
    \]
    The minimum value is \( -\frac{1}{4} (\beta - \alpha)^2 \).

The final answer is \( \boxed{ -\frac{1}{4} (\beta - \alpha)^2 } \)