Let $ A$ be an integer not equal to $ 0$. Solve the following system of equations in $ \mathbb{Z}^3$.
$ x \plus{} y^2 \plus{} z^3 \equal{} A$
$ \frac {1}{x} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^3} \equal{} \frac {1}{A}$
$ xy^2z^3 \equal{} A^2$
1. Given the system of equations:
   \[
   \begin{cases}
   x + y^2 + z^3 = A \\
   \frac{1}{x} + \frac{1}{y^2} + \frac{1}{z^3} = \frac{1}{A} \\
   xy^2z^3 = A^2
   \end{cases}
   \]
   where \( A \) is an integer not equal to 0.

2. From the second equation, multiply both sides by \( A \):
   \[
   A \left( \frac{1}{x} + \frac{1}{y^2} + \frac{1}{z^3} \right) = 1
   \]
   which simplifies to:
   \[
   \frac{A}{x} + \frac{A}{y^2} + \frac{A}{z^3} = 1
   \]

3. From the third equation, we have:
   \[
   xy^2z^3 = A^2
   \]

4. Consider the polynomial whose roots are \( x, y^2, z^3 \):
   \[
   t^3 - At^2 + At - A^2 = 0
   \]
   This polynomial is derived from the fact that the sum of the roots is \( A \), the sum of the product of the roots taken two at a time is \( A \), and the product of the roots is \( A^2 \).

5. Factorize the polynomial:
   \[
   (t - A)(t - (\sqrt{-A}))(t + (\sqrt{-A})) = 0
   \]
   This gives the roots:
   \[
   t = A, \quad t = \sqrt{-A}, \quad t = -\sqrt{-A}
   \]

6. Therefore, the possible values for \( x, y^2, z^3 \) are:
   \[
   \{ x, y^2, z^3 \} = \{ A, \sqrt{-A}, -\sqrt{-A} \}
   \]

7. Since \( A \) must be negative for \( \sqrt{-A} \) to be real, let \( A = -k^{12} \) for some integer \( k \). Then:
   \[
   y^2 = k^6, \quad z^3 = -k^4
   \]

8. The possible solutions for \( (x, y, z) \) are:
   \[
   (x, y, z) = (-k^{12}, \pm k^3, -k^2) \quad \text{or} \quad (x, y, z) = (-k^3, \pm k^3, -k^4)
   \]

The final answer is \( \boxed{ (x, y, z) = (-k^{12}, \pm k^3, -k^2) \quad \text{or} \quad (x, y, z) = (-k^3, \pm k^3, -k^4) } \)