Find all polynomials $P(x)$ which have the properties:
1) $P(x)$ is not a constant polynomial and is a mononic polynomial.
2) $P(x)$ has all real roots and no duplicate roots.
3) If $P(a)=0$ then $P(a|a|)=0$

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To find all polynomials \( P(x) \) that satisfy the given properties, we will analyze each condition step-by-step.

1. **Monic Polynomial and Non-constant**:
   - A monic polynomial is a polynomial whose leading coefficient is 1.
   - The polynomial is not constant, so it must have a degree of at least 1.

2. **All Real Roots and No Duplicate Roots**:
   - The polynomial must have distinct real roots.

3. **Condition on Roots**:
   - If \( P(a) = 0 \), then \( P(a|a|) = 0 \).

Let's analyze the third condition in detail. If \( P(a) = 0 \), then \( a \) is a root of \( P(x) \). The condition \( P(a|a|) = 0 \) implies that \( a|a| \) must also be a root of \( P(x) \).

### Case Analysis Based on \( a \):

- **Case 1: \( a = 0 \)**
  - If \( a = 0 \), then \( a|a| = 0 \). This condition is trivially satisfied.

- **Case 2: \( a > 0 \)**
  - If \( a > 0 \), then \( a|a| = a^2 \). Therefore, if \( a \) is a root, \( a^2 \) must also be a root.

- **Case 3: \( a < 0 \)**
  - If \( a < 0 \), then \( a|a| = a(-a) = -a^2 \). Therefore, if \( a \) is a root, \( -a^2 \) must also be a root.

Given that \( P(x) \) has distinct real roots, we need to find polynomials that satisfy these conditions. Let's consider the possible roots:

- If \( a = 1 \), then \( a|a| = 1 \).
- If \( a = -1 \), then \( a|a| = 1 \).

Thus, the roots must be such that \( a|a| \) is also a root. This implies that the roots must be \( 0 \), \( 1 \), and \(-1\).

### Constructing the Polynomial:
Given the roots \( 0 \), \( 1 \), and \(-1\), we can construct the polynomial as follows:

\[ P(x) = x(x-1)(x+1) \]

This polynomial is monic, has all real roots, no duplicate roots, and satisfies the condition \( P(a|a|) = 0 \) for any root \( a \).

### Verification:
- \( P(0) = 0 \)
- \( P(1) = 0 \)
- \( P(-1) = 0 \)

For \( a = 0 \):
\[ P(0|0|) = P(0) = 0 \]

For \( a = 1 \):
\[ P(1|1|) = P(1) = 0 \]

For \( a = -1 \):
\[ P(-1|-1|) = P(1) = 0 \]

Thus, the polynomial \( P(x) = x(x-1)(x+1) \) satisfies all the given conditions.

The final answer is \( \boxed{ P(x) = x(x-1)(x+1) } \).