Let $x<y$ be positive integers and $P=\frac{x^3-y}{1+xy}$. Find all integer values that $P$ can take.
1. We start with the given expression \( P = \frac{x^3 - y}{1 + xy} \) and need to find all integer values that \( P \) can take, given \( x < y \) are positive integers.

2. Consider the case where \( x = p^3 \) and \( y = p^5 - p \). Substituting these into the expression for \( P \):

   \[
   P = \frac{(p^3)^3 - (p^5 - p)}{1 + (p^3)(p^5 - p)}
   \]

3. Simplify the numerator and the denominator:

   \[
   P = \frac{p^9 - p^5 + p}{1 + p^8 - p^4}
   \]

4. Factor out \( p \) from the numerator and the denominator:

   \[
   P = \frac{p(p^8 - p^4 + 1)}{1 + p^8 - p^4}
   \]

5. Notice that the terms \( p^8 - p^4 + 1 \) in the numerator and the denominator cancel out:

   \[
   P = p
   \]

6. For \( p \geq 2 \), we have \( x = p^3 \) and \( y = p^5 - p \), which ensures \( x < y \) since \( p^3 < p^5 - p \) for \( p \geq 2 \).

7. Check the case \( p = 0 \):

   \[
   x = 2, \quad y = 8
   \]

   Substituting these values into the expression for \( P \):

   \[
   P = \frac{2^3 - 8}{1 + 2 \cdot 8} = \frac{8 - 8}{1 + 16} = \frac{0}{17} = 0
   \]

   Thus, \( p = 0 \) is also a valid solution.

8. Consider the case \( p \leq -1 \):

   \[
   y - x^3 \geq 1 + xy \implies x > xy
   \]

   This leads to a contradiction since \( x \) and \( y \) are positive integers.

9. For \( p = 1 \):

   \[
   x^3 - 1 = (x + 1)y \implies x + 1 \mid x^3 - 1
   \]

   Since \( x + 1 \mid x^3 + 1 \), we have \( x + 1 \mid 2 \). The possible values for \( x + 1 \) are \( \pm 1, \pm 2 \). Since \( x \) is positive, \( x + 1 = 2 \implies x = 1 \). This gives \( y = 0 \), which contradicts \( y \) being positive.

10. Therefore, the only valid values for \( p \) are \( p \geq 2 \) and \( p = 0 \).

The final answer is \( \boxed{ p \geq 2 } \) and \( p = 0 \).