What is the sum of all distinct values of $x$ that satisfy $x^4-x^3-7x^2+13x-6=0$?

[i]2018 CCA Math Bonanza Lightning Round #1.4[/i]
1. **Identify the polynomial and the goal**: We are given the polynomial \( P(x) = x^4 - x^3 - 7x^2 + 13x - 6 \) and need to find the sum of all distinct values of \( x \) that satisfy \( P(x) = 0 \).

2. **Test simple roots**: We start by testing \( x = 1 \) and \( x = -1 \) to see if they are roots of the polynomial.
   - For \( x = 1 \):
     \[
     P(1) = 1^4 - 1^3 - 7 \cdot 1^2 + 13 \cdot 1 - 6 = 1 - 1 - 7 + 13 - 6 = 0
     \]
     So, \( x = 1 \) is a root.

3. **Factor out \( (x-1) \)**: Since \( x = 1 \) is a root, we can factor \( P(x) \) as \( (x-1)Q(x) \), where \( Q(x) \) is a cubic polynomial. Perform polynomial division to find \( Q(x) \):
   \[
   x^4 - x^3 - 7x^2 + 13x - 6 \div (x-1) = x^3 - 7x + 6
   \]
   So, \( P(x) = (x-1)(x^3 - 7x + 6) \).

4. **Test \( x = 1 \) again**: Check if \( x = 1 \) is a root of \( x^3 - 7x + 6 \):
   \[
   1^3 - 7 \cdot 1 + 6 = 1 - 7 + 6 = 0
   \]
   So, \( x = 1 \) is a root again.

5. **Factor out \( (x-1) \) again**: Since \( x = 1 \) is a root of \( x^3 - 7x + 6 \), we can factor it further:
   \[
   x^3 - 7x + 6 \div (x-1) = x^2 + x - 6
   \]
   So, \( P(x) = (x-1)^2 (x^2 + x - 6) \).

6. **Factor the quadratic**: Factor \( x^2 + x - 6 \):
   \[
   x^2 + x - 6 = (x-2)(x+3)
   \]
   So, \( P(x) = (x-1)^2 (x-2)(x+3) \).

7. **Identify distinct roots**: The distinct roots of \( P(x) = 0 \) are \( x = 1, 2, -3 \).

8. **Sum of distinct roots**: Add the distinct roots:
   \[
   1 + 2 + (-3) = 0
   \]

The final answer is \(\boxed{0}\)