Find all positive integers $ x,y,z,t$ such that $ x,y,z$ are pairwise coprime and $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} xyzt$.
To find all positive integers \( x, y, z, t \) such that \( x, y, z \) are pairwise coprime and \((x + y)(y + z)(z + x) = xyzt\), we start by analyzing the given equation.

1. **Initial Analysis:**
   \[
   (x + y)(y + z)(z + x) = xyzt
   \]
   Since \( x, y, z \) are pairwise coprime, we can infer that \( x, y, z \) must divide the left-hand side of the equation.

2. **Divisibility Conditions:**
   \[
   xyz \mid (x + y)(y + z)(z + x)
   \]
   This implies:
   \[
   \begin{cases}
   x \mid (x + y)(y + z)(z + x) \\
   y \mid (x + y)(y + z)(z + x) \\
   z \mid (x + y)(y + z)(z + x)
   \end{cases}
   \]
   Since \( x, y, z \) are pairwise coprime, we can simplify this to:
   \[
   \begin{cases}
   x \mid y + z \\
   y \mid z + x \\
   z \mid x + y
   \end{cases}
   \]

3. **Summing the Divisibility Conditions:**
   \[
   xyz \mid x + y + z
   \]
   Without loss of generality, assume \( x \leq y \leq z \). Then:
   \[
   xyz \leq x + y + z \leq 3z
   \]
   This implies:
   \[
   x = 1
   \]

4. **Substituting \( x = 1 \):**
   \[
   (1 + y)(y + z)(z + 1) = 1 \cdot y \cdot z \cdot t
   \]
   Simplifying:
   \[
   (1 + y)(y + z)(z + 1) = yzt
   \]

5. **Analyzing \( y \) and \( z \):**
   \[
   yz \mid 1 + y + z
   \]
   Since \( yz \leq 1 + y + z \), we have:
   \[
   yz \leq 1 + y + z
   \]
   Rearranging:
   \[
   yz - y - z \leq 1
   \]
   Adding 1 to both sides:
   \[
   (y - 1)(z - 1) \leq 2
   \]

6. **Possible Values for \( y \) and \( z \):**
   - If \( y = 1 \):
     \[
     z \mid 2 + z \implies z = 1 \text{ or } 2
     \]
     - For \( z = 1 \):
       \[
       (x, y, z, t) = (1, 1, 1, 8)
       \]
     - For \( z = 2 \):
       \[
       (x, y, z, t) = (1, 1, 2, 9)
       \]

   - If \( y = 2 \):
     \[
     2z \mid 3 + z \implies z = 3
     \]
     - For \( z = 3 \):
       \[
       (x, y, z, t) = (1, 2, 3, 10)
       \]

7. **Conclusion:**
   The solutions are:
   \[
   (x, y, z, t) = (1, 1, 1, 8), (1, 1, 2, 9), (1, 2, 3, 10)
   \]

The final answer is \( \boxed{ (1, 1, 1, 8), (1, 1, 2, 9), (1, 2, 3, 10) } \)