Suppose $x^9 = 1$ but $x^3 \ne 1$. Find a polynomial of minimal degree equal to $\frac{1}{1+x}$ .
1. Given \( x^9 = 1 \) and \( x^3 \ne 1 \), we start by factorizing \( x^9 - 1 = 0 \). This can be written as:
   \[
   x^9 - 1 = (x^3)^3 - 1^3 = (x^3 - 1)(x^6 + x^3 + 1) = 0
   \]
   Since \( x^3 \ne 1 \), it follows that \( x^3 - 1 \ne 0 \). Therefore, we must have:
   \[
   x^6 + x^3 + 1 = 0
   \]

2. We need to find a polynomial of minimal degree equal to \( \frac{1}{1+x} \). Using the equation \( x^6 + x^3 + 1 = 0 \), we can express \( x^6 + x^3 \) as:
   \[
   x^6 + x^3 = -1
   \]

3. We transform \( \frac{1}{1+x} \) using the above result:
   \[
   \frac{1}{1+x} = \frac{x^6 + x^3}{x+1}
   \]

4. To simplify \( \frac{x^6 + x^3}{x+1} \), we perform polynomial long division:
   \[
   \frac{x^6 + x^3}{x+1} = x^5 - x^4 + x^3
   \]

5. Therefore, the minimal degree polynomial that is equal to \( \frac{1}{1+x} \) is:
   \[
   x^5 - x^4 + x^3
   \]

The final answer is \( \boxed{ x^5 - x^4 + x^3 } \).