Suppose that the rational numbers $a, b$ and $c$ are the roots of the equation $x^3+ax^2 + bx + c = 0$. Find all such rational numbers $a, b$ and $c$. Justify your answer
1. Given the polynomial \( x^3 + ax^2 + bx + c = 0 \) with rational roots \( a, b, \) and \( c \), we need to find all such rational numbers \( a, b, \) and \( c \).

2. By Vieta's formulas, for a polynomial \( x^3 + px^2 + qx + r = 0 \) with roots \( \alpha, \beta, \gamma \), we have:
   \[
   \alpha + \beta + \gamma = -p,
   \]
   \[
   \alpha\beta + \beta\gamma + \gamma\alpha = q,
   \]
   \[
   \alpha\beta\gamma = -r.
   \]

3. Applying Vieta's formulas to our polynomial \( x^3 + ax^2 + bx + c = 0 \), we get:
   \[
   \alpha + \beta + \gamma = -a,
   \]
   \[
   \alpha\beta + \beta\gamma + \gamma\alpha = b,
   \]
   \[
   \alpha\beta\gamma = -c.
   \]

4. Since \( \alpha, \beta, \gamma \) are rational, and the polynomial is monic (leading coefficient is 1), the roots must be integers (by the Rational Root Theorem).

5. Let's denote the roots by \( \alpha, \beta, \gamma \). Then the polynomial can be written as:
   \[
   (x - \alpha)(x - \beta)(x - \gamma) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma.
   \]

6. Comparing coefficients, we have:
   \[
   a = -(\alpha + \beta + \gamma),
   \]
   \[
   b = \alpha\beta + \beta\gamma + \gamma\alpha,
   \]
   \[
   c = -\alpha\beta\gamma.
   \]

7. We need to find all possible sets of integers \( \alpha, \beta, \gamma \) such that \( a, b, \) and \( c \) are rational numbers.

8. Consider the case where \( c = 0 \):
   \[
   \alpha\beta\gamma = 0 \implies \text{at least one of } \alpha, \beta, \gamma \text{ is zero}.
   \]
   Without loss of generality, let \( \gamma = 0 \). Then:
   \[
   \alpha + \beta = -a,
   \]
   \[
   \alpha\beta = b.
   \]
   Solving for \( \alpha \) and \( \beta \):
   \[
   \alpha + \beta = -a,
   \]
   \[
   \alpha\beta = b.
   \]
   Possible integer solutions for \( \alpha \) and \( \beta \) must satisfy these equations.

9. Consider the case where \( c \neq 0 \):
   \[
   \alpha\beta\gamma = -c \implies \alpha, \beta, \gamma \neq 0.
   \]
   We need to find integer solutions for \( \alpha, \beta, \gamma \) such that:
   \[
   \alpha + \beta + \gamma = -a,
   \]
   \[
   \alpha\beta + \beta\gamma + \gamma\alpha = b,
   \]
   \[
   \alpha\beta\gamma = -c.
   \]

10. Let's solve for specific values:
    - If \( \alpha = 1, \beta = -1, \gamma = -1 \):
      \[
      \alpha + \beta + \gamma = 1 - 1 - 1 = -1 \implies a = 1,
      \]
      \[
      \alpha\beta + \beta\gamma + \gamma\alpha = 1(-1) + (-1)(-1) + (-1)(1) = -1 + 1 - 1 = -1 \implies b = -1,
      \]
      \[
      \alpha\beta\gamma = 1(-1)(-1) = 1 \implies c = -1.
      \]
      This gives the solution \( (a, b, c) = (1, -1, -1) \).

    - If \( \alpha = 0, \beta = 0, \gamma = 0 \):
      \[
      \alpha + \beta + \gamma = 0 \implies a = 0,
      \]
      \[
      \alpha\beta + \beta\gamma + \gamma\alpha = 0 \implies b = 0,
      \]
      \[
      \alpha\beta\gamma = 0 \implies c = 0.
      \]
      This gives the solution \( (a, b, c) = (0, 0, 0) \).

    - If \( \alpha = 1, \beta = -2, \gamma = 0 \):
      \[
      \alpha + \beta + \gamma = 1 - 2 + 0 = -1 \implies a = 1,
      \]
      \[
      \alpha\beta + \beta\gamma + \gamma\alpha = 1(-2) + (-2)(0) + (0)(1) = -2 \implies b = -2,
      \]
      \[
      \alpha\beta\gamma = 1(-2)(0) = 0 \implies c = 0.
      \]
      This gives the solution \( (a, b, c) = (1, -2, 0) \).

11. Therefore, the possible sets of rational numbers \( (a, b, c) \) are:
    \[
    (1, -1, -1), (0, 0, 0), (1, -2, 0).
    \]

The final answer is \( \boxed{ (1, -1, -1), (0, 0, 0), (1, -2, 0) } \).