Let $F = \max_{1 \leq x \leq 3} |x^3 - ax^2 - bx - c|$.  When $a$, $b$, $c$ run over all the real numbers, find the smallest possible value of $F$.
To solve the problem, we need to find the smallest possible value of \( F = \max_{1 \leq x \leq 3} |x^3 - ax^2 - bx - c| \) when \( a, b, c \) run over all real numbers. 

1. **Substitution and Transformation:**
   We start by substituting \( y = x - 2 \). This transforms the interval \( 1 \leq x \leq 3 \) to \( -1 \leq y \leq 1 \). The polynomial \( x^3 - ax^2 - bx - c \) becomes:
   \[
   (y+2)^3 - a(y+2)^2 - b(y+2) - c
   \]
   Expanding and simplifying, we get:
   \[
   y^3 + (6 - a)y^2 + (12 - 2a - b)y + (8 - 4a - 2b - c)
   \]
   Let \( a' = 6 - a \), \( b' = 12 - 2a - b \), and \( c' = 8 - 4a - 2b - c \). The problem reduces to finding the minimum value of:
   \[
   G = \max_{-1 \leq y \leq 1} |y^3 + a'y^2 + b'y + c'|
   \]
   over all real numbers \( a', b', c' \).

2. **Form of the Polynomial:**
   We need to minimize the maximum absolute value of the polynomial \( y^3 + a'y^2 + b'y + c' \) over the interval \( -1 \leq y \leq 1 \). 

3. **Choosing Coefficients:**
   Consider the polynomial \( y^3 - \frac{3}{4}y \). We need to check if this polynomial achieves the minimum possible maximum value over the interval \( -1 \leq y \leq 1 \). 

4. **Checking Extremal Values:**
   The extremal values of \( y^3 - \frac{3}{4}y \) occur at \( y = \pm 1 \) and \( y = 0 \). Evaluating at these points:
   \[
   y = 1: \quad 1^3 - \frac{3}{4}(1) = \frac{1}{4}
   \]
   \[
   y = -1: \quad (-1)^3 - \frac{3}{4}(-1) = -\frac{1}{4}
   \]
   \[
   y = 0: \quad 0^3 - \frac{3}{4}(0) = 0
   \]
   The maximum absolute value is \( \frac{1}{4} \).

5. **Verification:**
   Suppose there exists a polynomial \( y^3 + a'y^2 + b'y + c' \) with a smaller maximum absolute value over \( -1 \leq y \leq 1 \). By differentiation, the extremal values of \( y^3 + b'y \) occur at \( y = \pm \sqrt{-b'/3} \) (if \( b' < 0 \)) and \( y = \pm 1 \). 

   If \( b' < -\frac{3}{4} \), then \( y = 1 \) gives a maximum greater than \( \frac{1}{4} \), and \( y = -1 \) gives a minimum less than \( -\frac{1}{4} \). This implies that the magnitude of the polynomial exceeds \( \frac{1}{4} \), which is inadmissible.

   If \( b' > -\frac{3}{4} \), similar arguments show that the magnitude of the polynomial exceeds \( \frac{1}{4} \).

6. **Conclusion:**
   Therefore, the smallest possible value of \( F \) is achieved when \( a = 0 \), \( b = -\frac{3}{4} \), and \( c = 0 \), giving \( F = \frac{1}{4} \).

The final answer is \( \boxed{\frac{1}{4}} \).