The polynomials $P_{n}(x)$ are defined by $P_{0}(x)=0,P_{1}(x)=x$ and \[P_{n}(x)=xP_{n-1}(x)+(1-x)P_{n-2}(x) \quad n\geq 2\] For every natural number $n\geq 1$, find all real numbers $x$ satisfying the equation $P_{n}(x)=0$.
1. **Define the polynomials and recurrence relation:**
   The polynomials \( P_n(x) \) are defined by:
   \[
   P_0(x) = 0, \quad P_1(x) = x
   \]
   and for \( n \geq 2 \):
   \[
   P_n(x) = xP_{n-1}(x) + (1-x)P_{n-2}(x)
   \]

2. **Find the general form of \( P_n(x) \):**
   We use the recurrence relation to find a pattern or closed form for \( P_n(x) \). By solving the recurrence relation, we get:
   \[
   P_n(x) = \frac{x((x-1)^n - 1)}{x-2}
   \]
   This can be derived using standard techniques for solving linear recurrence relations with constant coefficients.

3. **Identify the roots of \( P_n(x) \):**
   To find the roots of \( P_n(x) = 0 \), we need to solve:
   \[
   \frac{x((x-1)^n - 1)}{x-2} = 0
   \]
   The fraction is zero when the numerator is zero, provided the denominator is not zero. Thus, we solve:
   \[
   x((x-1)^n - 1) = 0
   \]

4. **Solve the equation \( x((x-1)^n - 1) = 0 \):**
   This equation has two factors:
   \[
   x = 0 \quad \text{or} \quad (x-1)^n = 1
   \]
   - For \( x = 0 \), this is a root.
   - For \( (x-1)^n = 1 \), we need to consider the values of \( x \):
     \[
     (x-1)^n = 1 \implies x-1 = 1 \quad \text{or} \quad x-1 = -1
     \]
     - If \( x-1 = 1 \), then \( x = 2 \).
     - If \( x-1 = -1 \), then \( x = 0 \) (but this is already considered).

5. **Check the validity of \( x = 2 \):**
   We need to ensure that \( x = 2 \) does not make the denominator zero:
   \[
   \frac{x((x-1)^n - 1)}{x-2} \quad \text{is undefined for} \quad x = 2
   \]
   Therefore, \( x = 2 \) is not a valid root.

6. **Conclusion:**
   The only real root of \( P_n(x) = 0 \) for every natural number \( n \geq 1 \) is \( x = 0 \).

The final answer is \( \boxed{ x = 0 } \).