Find all  polynomials $P(x)$ with real coefficents such that \[P(P(x))=(x^2+x+1)\cdot P(x)\]   where $x \in \mathbb{R}$
1. **Constant Polynomial Case:**
   - First, consider the case where \( P(x) \) is a constant polynomial. Let \( P(x) = c \) where \( c \) is a constant.
   - Substituting \( P(x) = c \) into the given equation \( P(P(x)) = (x^2 + x + 1) \cdot P(x) \), we get:
     \[
     P(c) = (x^2 + x + 1) \cdot c
     \]
   - Since \( P(x) = c \), we have \( P(c) = c \). Therefore, the equation becomes:
     \[
     c = (x^2 + x + 1) \cdot c
     \]
   - For this to hold for all \( x \in \mathbb{R} \), we must have \( c = 0 \). Thus, the only constant solution is:
     \[
     \boxed{P(x) = 0 \text{ for all } x}
     \]

2. **Non-Constant Polynomial Case:**
   - Let \( P(x) \) be a non-constant polynomial of degree \( n \). Then \( P(P(x)) \) has degree \( n^2 \).
   - The right-hand side of the given equation is \( (x^2 + x + 1) \cdot P(x) \), which has degree \( 2 + n \).
   - Equating the degrees, we get:
     \[
     n^2 = n + 2
     \]
   - Solving this quadratic equation:
     \[
     n^2 - n - 2 = 0
     \]
     \[
     (n - 2)(n + 1) = 0
     \]
     \[
     n = 2 \text{ or } n = -1
     \]
   - Since \( n \) must be a non-negative integer, we have \( n = 2 \).

3. **Form of the Polynomial:**
   - Let \( P(x) = ax^2 + bx + c \). Since \( P(x) \) is a polynomial of degree 2, we can write:
     \[
     P(x) = ax^2 + bx + c
     \]
   - Substituting \( P(x) \) into the given equation:
     \[
     P(P(x)) = (x^2 + x + 1) \cdot P(x)
     \]
   - We need to find \( P(P(x)) \):
     \[
     P(P(x)) = P(ax^2 + bx + c) = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c
     \]
   - Expanding \( (ax^2 + bx + c)^2 \):
     \[
     (ax^2 + bx + c)^2 = a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2
     \]
   - Therefore:
     \[
     P(P(x)) = a(a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2) + b(ax^2 + bx + c) + c
     \]
     \[
     = a^3x^4 + 2a^2bx^3 + a(2ac + b^2)x^2 + 2abcx + ac^2 + abx^2 + b^2x + bc + c
     \]
   - Simplifying and collecting like terms:
     \[
     P(P(x)) = a^3x^4 + 2a^2bx^3 + (a(2ac + b^2) + ab)x^2 + (2abc + b^2)x + (ac^2 + bc + c)
     \]
   - On the other hand, the right-hand side of the given equation is:
     \[
     (x^2 + x + 1) \cdot (ax^2 + bx + c) = ax^4 + (a + b)x^3 + (a + b + c)x^2 + (b + c)x + c
     \]
   - Equating the coefficients of \( x^4, x^3, x^2, x \), and the constant term from both sides, we get:
     \[
     a^3 = a \quad \Rightarrow \quad a(a^2 - 1) = 0 \quad \Rightarrow \quad a = 1 \text{ or } a = -1
     \]
     \[
     2a^2b = a + b \quad \Rightarrow \quad 2b = 1 + b \quad \Rightarrow \quad b = 1
     \]
     \[
     a(2ac + b^2) + ab = a + b + c \quad \Rightarrow \quad 2c + 1 + 1 = 1 + 1 + c \quad \Rightarrow \quad c = 0
     \]
   - Therefore, the polynomial \( P(x) \) is:
     \[
     P(x) = x^2 + x
     \]
   - Verifying, we substitute \( P(x) = x^2 + x \) back into the original equation:
     \[
     P(P(x)) = P(x^2 + x) = (x^2 + x)^2 + (x^2 + x) = x^4 + 2x^3 + x^2 + x^2 + x = x^4 + 2x^3 + 2x^2 + x
     \]
     \[
     (x^2 + x + 1) \cdot P(x) = (x^2 + x + 1) \cdot (x^2 + x) = x^4 + x^3 + x^2 + x^3 + x^2 + x = x^4 + 2x^3 + 2x^2 + x
     \]
   - Both sides match, confirming that \( P(x) = x^2 + x \) is indeed a solution.

Conclusion:
\[
\boxed{P(x) = 0 \text{ or } P(x) = x^2 + x}
\]