Find all polynomials $P(X)$ with real coefficients such that if real numbers $x,y$ and $z$ satisfy $x+y+z=0,$ then the points $\left(x,P(x)\right), \left(y,P(y)\right), \left(z,P(z)\right)$ are all colinear.
1. **Assume the form of the polynomial:**
   We need to find all polynomials \( P(x) \) such that if \( x + y + z = 0 \), then the points \((x, P(x))\), \((y, P(y))\), and \((z, P(z))\) are collinear. 

2. **Collinearity condition:**
   Three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are collinear if the area of the triangle they form is zero. This can be expressed using the determinant:
   \[
   \begin{vmatrix}
   x_1 & y_1 & 1 \\
   x_2 & y_2 & 1 \\
   x_3 & y_3 & 1
   \end{vmatrix} = 0
   \]
   For our points \((x, P(x))\), \((y, P(y))\), and \((z, P(z))\), this becomes:
   \[
   \begin{vmatrix}
   x & P(x) & 1 \\
   y & P(y) & 1 \\
   z & P(z) & 1
   \end{vmatrix} = 0
   \]

3. **Simplify the determinant:**
   Expanding the determinant, we get:
   \[
   x \left( P(y) - P(z) \right) - y \left( P(x) - P(z) \right) + z \left( P(x) - P(y) \right) = 0
   \]
   Since \( x + y + z = 0 \), we can substitute \( z = -x - y \):
   \[
   x \left( P(y) - P(-x-y) \right) - y \left( P(x) - P(-x-y) \right) + (-x-y) \left( P(x) - P(y) \right) = 0
   \]

4. **Analyze the polynomial:**
   To satisfy this equation for all \( x \) and \( y \), we need to find a polynomial \( P(x) \) such that the above equation holds. Let's assume \( P(x) \) is a polynomial of degree \( n \):
   \[
   P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
   \]
   Since \( P(x) \) must satisfy the collinearity condition for all \( x \) and \( y \), we need to check the degrees of the terms.

5. **Odd function property:**
   By shifting \( P \) downwards, we can assume without loss of generality that \( P(0) = 0 \). This implies that \( P(x) \) is an odd function because \( P(-x) = -P(x) \).

6. **Form of the polynomial:**
   Given that \( P(x) \) is an odd function, it can only contain odd powers of \( x \). Therefore, \( P(x) \) must be of the form:
   \[
   P(x) = a_1 x + a_3 x^3 + a_5 x^5 + \cdots
   \]
   However, we need to ensure that the collinearity condition is satisfied for all \( x \) and \( y \). The simplest polynomial that satisfies this condition is a cubic polynomial:
   \[
   P(x) = ax^3 + bx
   \]

7. **General solution:**
   Since we initially shifted \( P \) downwards, we need to add back the constant term \( c \). Therefore, the general form of the polynomial \( P(x) \) is:
   \[
   P(x) = ax^3 + bx + c
   \]

The final answer is \( \boxed{ P(x) = ax^3 + bx + c } \) where \( a, b, c \in \mathbb{R} \).