Find all polynomials $P(x)$ with real coefficents, such that for all $x,y,z$ satisfying $x+y+z=0$, the equation below is true: \[P(x+y)^3+P(y+z)^3+P(z+x)^3=3P((x+y)(y+z)(z+x))\]
1. **Substitute \( x = y = z = 0 \):**
   \[
   P(0+0)^3 + P(0+0)^3 + P(0+0)^3 = 3P((0+0)(0+0)(0+0))
   \]
   Simplifying, we get:
   \[
   3P(0)^3 = 3P(0)
   \]
   Dividing both sides by 3, we obtain:
   \[
   P(0)^3 = P(0)
   \]
   This implies:
   \[
   P(0) = 0 \quad \text{or} \quad P(0) = 1 \quad \text{or} \quad P(0) = -1
   \]

2. **Substitute \( x = y+z \):**
   \[
   P(-y)^3 + P(-z)^3 + P(y+z)^3 = 3P(yz(y+z))
   \]
   This must hold for all \( y \) and \( z \).

3. **Substitute \( y = -z \):**
   \[
   P(z)^3 + P(-z)^3 + P(0)^3 = 3P(0)
   \]
   Since \( P(0) = 0 \) from step 1, we have:
   \[
   P(z)^3 + P(-z)^3 = 0
   \]
   This implies:
   \[
   P(z)^3 = -P(-z)^3
   \]
   Since \( P(z) \) and \( P(-z) \) are real, we have:
   \[
   P(z) = -P(-z)
   \]
   Therefore, \( P \) is an odd function.

4. **Assume \( P(x) = a_n x^{2n+1} + \ldots + a_1 x \):**
   Since \( P \) is an odd function, it only contains odd powers of \( x \).

5. **Compare the highest degree terms:**
   On the left side of the original equation, the highest degree term is:
   \[
   (a_n (y+z)^{2n+1})^3 = a_n^3 (y+z)^{6n+3}
   \]
   On the right side, the highest degree term is:
   \[
   3a_n (yz(y+z))^{2n+1} = 3a_n (y^{2n+1} z^{2n+1} (y+z)^{2n+1})
   \]
   For the degrees to match, we must have:
   \[
   6n+3 = 6n+3
   \]
   This is always true, but we need to ensure the coefficients match.

6. **Match coefficients:**
   The term \( y^{6n+2} z \) appears on the left but not on the right if \( n > 0 \). Therefore, \( n = 0 \).

7. **Solve for \( a_n \):**
   If \( n = 0 \), then \( P(x) = a_1 x \). From the equation \( a_n^3 = a_n \), we get:
   \[
   a_1^3 = a_1
   \]
   This implies:
   \[
   a_1 = 0 \quad \text{or} \quad a_1 = 1 \quad \text{or} \quad a_1 = -1
   \]

Thus, the possible polynomials are:
\[
P(x) = 0, \quad P(x) = x, \quad P(x) = -x
\]

The final answer is \( \boxed{P(x) = 0, \quad P(x) = x, \quad P(x) = -x} \).