Find all cubic polynomials $x^3 +ax^2 +bx+c$ admitting the rational numbers $a$, $b$ and $c$ as roots.
To find all cubic polynomials \(x^3 + ax^2 + bx + c\) that admit the rational numbers \(a\), \(b\), and \(c\) as roots, we can use Vieta's formulas. Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.

Given the polynomial \(P(x) = x^3 + ax^2 + bx + c\) with roots \(a\), \(b\), and \(c\), Vieta's formulas give us the following relationships:
1. \(a + b + c = -a\)
2. \(ab + bc + ca = b\)
3. \(abc = -c\)

Let's analyze these equations step by step.

1. **Sum of the roots:**
   \[
   a + b + c = -a
   \]
   Simplifying, we get:
   \[
   2a + b + c = 0
   \]

2. **Product of the roots taken two at a time:**
   \[
   ab + bc + ca = b
   \]

3. **Product of the roots:**
   \[
   abc = -c
   \]

We will consider two cases: \(c = 0\) and \(c \neq 0\).

### Case 1: \(c = 0\)
If \(c = 0\), then the polynomial simplifies to \(P(x) = x^3 + ax^2 + bx\). The roots are \(a\), \(b\), and \(0\).

From Vieta's formulas:
1. \(2a + b = 0\)
2. \(ab = b\)

From the second equation, if \(b \neq 0\), we can divide both sides by \(b\) to get:
\[
a = 1
\]
Substituting \(a = 1\) into the first equation:
\[
2(1) + b = 0 \implies b = -2
\]
Thus, the polynomial is:
\[
P(x) = x^3 + x^2 - 2x
\]

If \(b = 0\), then from the first equation:
\[
2a = 0 \implies a = 0
\]
Thus, the polynomial is:
\[
P(x) = x^3
\]

### Case 2: \(c \neq 0\)
If \(c \neq 0\), then from \(abc = -c\), we get:
\[
ab = -1
\]
Thus, \(a\) and \(b\) are non-zero. We can multiply the second Vieta's formula by \(a\):
\[
a(ab + bc + ca) = ab \implies a^2b + abc + ca^2 = ab
\]
Substituting \(abc = -c\) and \(ab = -1\):
\[
a^2b - c + ca^2 = -1
\]
Simplifying, we get:
\[
a^2b + ca^2 - c = -1
\]
Rearranging, we get:
\[
c(a^2 + 1) = a - 1 \implies c = \frac{a - 1}{a^2 + 1}
\]

On the other hand, from \(a(2a + b + c) = 0\):
\[
a(-2a - c) = -1 \implies c = \frac{1 - 2a^2}{a}
\]

Equating the two expressions for \(c\):
\[
\frac{a - 1}{a^2 + 1} = \frac{1 - 2a^2}{a}
\]

Cross-multiplying, we get:
\[
a(a - 1) = (a^2 + 1)(1 - 2a^2)
\]
Expanding and simplifying:
\[
a^2 - a = 1 + a^2 - 2a^4 - 2a^2
\]
\[
-a = 1 - 2a^4 - a^2
\]
\[
2a^4 + a^2 - a - 1 = 0
\]

This is a quartic equation in \(a\). To check for rational roots, we use the Rational Root Theorem, which states that any rational root, in the form \(\frac{p}{q}\), must be a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is \(-1\) and the leading coefficient is \(2\). The possible rational roots are \(\pm 1, \pm \frac{1}{2}\).

Testing these values, we find that none of them satisfy the equation \(2a^4 + a^2 - a - 1 = 0\). Therefore, there are no rational solutions in this case.

Thus, the only polynomials are those found in Case 1.

The final answer is \(\boxed{\{x^3, x^3 + x^2 - 2x\}}\).