Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $$x^3+y^3=p(xy+p).$$
1. **Initial Setup and Assumptions:**
   We start with the equation:
   \[
   x^3 + y^3 = p(xy + p)
   \]
   where \( p \) is a prime number and \( x \) and \( y \) are positive integers. We need to determine all possible values of \( x \), \( y \), and \( p \).

2. **Prime Divisibility Argument:**
   If \( p \) divides either \( x \) or \( y \), then \( p \) must divide both \( x \) and \( y \). This would imply that the left-hand side (LHS) of the equation \( x^3 + y^3 \) is divisible by \( p^3 \), while the right-hand side (RHS) \( p(xy + p) \) is only divisible by \( p^2 \), leading to a contradiction. Therefore, \( p \) does not divide \( x \) or \( y \).

3. **GCD Argument:**
   If another prime \( q \) divides both \( x \) and \( y \), then taking the equation modulo \( q \) gives \( q \mid p^2 \), implying \( q = p \), which is a contradiction. Thus, \( \gcd(x, y) = 1 \).

4. **Factoring the Equation:**
   We rewrite the equation as:
   \[
   (x + y)(x^2 - xy + y^2) = p(xy + p)
   \]
   By the Four Number Lemma, there exist positive integers \( a, b, c, d \) such that:
   \[
   ab = x + y, \quad cd = x^2 - xy + y^2, \quad ac = p, \quad bd = xy + p
   \]
   Since \( p \) is a prime, we have two cases: \( a = p, c = 1 \) or \( a = 1, c = p \).

5. **Case 1: \( a = p \) and \( c = 1 \):**
   - We get:
     \[
     x + y = pb, \quad x^2 - xy + y^2 = d, \quad xy + p = bd
     \]
   - By AM-GM inequality, \( x^2 + y^2 \geq 2xy \), so \( d \geq xy \). Using the third equation, we get \( p \geq (b-1)xy \). Therefore:
     \[
     x + y \geq b(b-1)xy
     \]
     This is impossible if \( b > 1 \) (since there are no solutions when \( x = y = 1 \)). Thus, \( b = 1 \), so \( p = x + y \). Dividing both sides of the equation by \( x + y \), we get:
     \[
     x^2 - xy + y^2 = xy + p \implies p = (x - y)^2
     \]
     This is absurd since no prime is a perfect square. Therefore, this case has no solutions.

6. **Case 2: \( a = 1 \) and \( c = p \):**
   - We get:
     \[
     x + y = b, \quad x^2 - xy + y^2 = pd, \quad xy + p = bd
     \]

7. **Claim: \( d = 3 \):**
   - Suppose \( d \neq 3 \). Notice that:
     \[
     (x + y)^2 - 3xy = x^2 - xy + y^2
     \]
     So:
     \[
     b^2 - 3(bd - p) = pd \implies b^2 - 3bd = pd - 3p
     \]
     Thus, \( b \mid pd - 3p = p(d - 3) \). If \( p \mid b \), then taking the third equation modulo \( p \) gives \( p \mid xy \), which we know is not true. Thus, \( \gcd(b, p) = 1 \), so \( b \mid d - 3 \).

   - If \( d < 3 \), then \( b = 2 \). This case leads to no solutions, so assume \( d > 3 \). We have \( d - 3 \geq b \), so \( d > b = x + y \).

   - Now we have:
     \[
     d = \frac{xy + p}{x + y} \leq \frac{xy + pd}{x + y} = \frac{x^2 + y^2}{x + y} \leq x + y = b
     \]
     This is absurd since \( d > b \). Therefore, \( d = 3 \).

8. **Final Calculation:**
   - Now, we have:
     \[
     x^2 - xy + y^2 = b^2 - 3(3b - p) = b^2 - 9b + 3p
     \]
     Since this is equal to \( 3p \), we get:
     \[
     b^2 - 9b = 0 \implies b = x + y = 9
     \]
   - Since \( \gcd(x, y) = 1 \), we must have \( \{x, y\} = \{1, 8\}, \{2, 7\}, \{4, 5\} \), which gives the desired solution set.

The final answer is \(\boxed{(x, y, p) = (8, 1, 19), (1, 8, 19), (7, 2, 13), (2, 7, 13), (5, 4, 7), (4, 5, 7)}\).