Solve the system of equations $ \begin{cases} x^3 \plus{} x(y \minus{} z)^2 \equal{} 2\\ y^3 \plus{} y(z \minus{} x)^2 \equal{} 30\\ z^3 \plus{} z(x \minus{} y)^2 \equal{} 16\end{cases}$.
We are given the system of equations:
\[
\begin{cases}
x^3 + x(y - z)^2 = 2 \\
y^3 + y(z - x)^2 = 30 \\
z^3 + z(x - y)^2 = 16
\end{cases}
\]

1. **Rewrite each equation in a more convenient form:**

   \[
   x^3 + x(y - z)^2 = 2
   \]
   \[
   y^3 + y(z - x)^2 = 30
   \]
   \[
   z^3 + z(x - y)^2 = 16
   \]

2. **Combine the equations:**

   Add the first two equations and subtract twice the third equation:
   \[
   (x^3 + x(y - z)^2) + (y^3 + y(z - x)^2) - 2(z^3 + z(x - y)^2) = 2 + 30 - 2 \cdot 16
   \]
   Simplify the right-hand side:
   \[
   2 + 30 - 32 = 0
   \]

3. **Factor the left-hand side:**

   Notice that the left-hand side can be factored as:
   \[
   (x + y - 2z)(x^2 + y^2 + z^2) = 0
   \]
   This gives us two possible cases:
   \[
   x + y - 2z = 0 \quad \text{or} \quad x^2 + y^2 + z^2 = 0
   \]

4. **Analyze the cases:**

   - **Case 1: \( x + y - 2z = 0 \)**

     Solving for \( z \):
     \[
     z = \frac{x + y}{2}
     \]

     Substitute \( z = \frac{x + y}{2} \) into the original equations:
     \[
     \begin{cases}
     x^3 + x(y - \frac{x + y}{2})^2 = 2 \\
     y^3 + y(\frac{x + y}{2} - x)^2 = 30 \\
     (\frac{x + y}{2})^3 + \frac{x + y}{2}(x - y)^2 = 16
     \end{cases}
     \]

     Simplify each equation:
     \[
     x^3 + x\left(\frac{2y - x - y}{2}\right)^2 = 2 \implies x^3 + x\left(\frac{y - x}{2}\right)^2 = 2 \implies x^3 + \frac{x(y - x)^2}{4} = 2
     \]
     \[
     y^3 + y\left(\frac{x + y - 2x}{2}\right)^2 = 30 \implies y^3 + y\left(\frac{y - x}{2}\right)^2 = 30 \implies y^3 + \frac{y(y - x)^2}{4} = 30
     \]
     \[
     \left(\frac{x + y}{2}\right)^3 + \frac{x + y}{2}(x - y)^2 = 16 \implies \frac{(x + y)^3}{8} + \frac{(x + y)(x - y)^2}{2} = 16
     \]

     Let \( d = y - x \), then \( y = x + d \) and \( z = \frac{x + (x + d)}{2} = x + \frac{d}{2} \).

     Substitute \( y = x + d \) and \( z = x + \frac{d}{2} \) into the equations:
     \[
     x^3 + x\left(\frac{d}{2}\right)^2 = 2 \implies x^3 + \frac{xd^2}{4} = 2
     \]
     \[
     (x + d)^3 + (x + d)\left(\frac{d}{2}\right)^2 = 30 \implies (x + d)^3 + \frac{(x + d)d^2}{4} = 30
     \]
     \[
     \left(x + \frac{d}{2}\right)^3 + \left(x + \frac{d}{2}\right)(d)^2 = 16 \implies \left(x + \frac{d}{2}\right)^3 + (x + \frac{d}{2})d^2 = 16
     \]

     Simplify each equation:
     \[
     x^3 + \frac{xd^2}{4} = 2
     \]
     \[
     (x + d)^3 + \frac{(x + d)d^2}{4} = 30
     \]
     \[
     \left(x + \frac{d}{2}\right)^3 + (x + \frac{d}{2})d^2 = 16
     \]

     Solve the simplified equations:
     \[
     x^3 + \frac{xd^2}{4} = 2
     \]
     \[
     (x + d)^3 + \frac{(x + d)d^2}{4} = 30
     \]
     \[
     \left(x + \frac{d}{2}\right)^3 + (x + \frac{d}{2})d^2 = 16
     \]

     Subtract the first equation from the second:
     \[
     (x + d)^3 + \frac{(x + d)d^2}{4} - x^3 - \frac{xd^2}{4} = 30 - 2
     \]
     Simplify:
     \[
     (x^3 + 3x^2d + 3xd^2 + d^3) + \frac{xd^2 + 2xd^2 + d^3}{4} - x^3 - \frac{xd^2}{4} = 28
     \]
     \[
     3x^2d + 3xd^2 + d^3 + \frac{3xd^2 + d^3}{4} = 28
     \]
     \[
     3x^2d + 3xd^2 + d^3 + \frac{3xd^2 + d^3}{4} = 28
     \]

     Subtract the first equation from the third:
     \[
     \left(x + \frac{d}{2}\right)^3 + (x + \frac{d}{2})d^2 - x^3 - \frac{xd^2}{4} = 16 - 2
     \]
     Simplify:
     \[
     \left(x^3 + \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8}\right) + \left(xd^2 + \frac{d^3}{2}\right) - x^3 - \frac{xd^2}{4} = 14
     \]
     \[
     \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8} + xd^2 + \frac{d^3}{2} - \frac{xd^2}{4} = 14
     \]
     \[
     \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8} + xd^2 + \frac{d^3}{2} - \frac{xd^2}{4} = 14
     \]

     Subtract the second equation from the third:
     \[
     \left(x + \frac{d}{2}\right)^3 + (x + \frac{d}{2})d^2 - (x + d)^3 - \frac{(x + d)d^2}{4} = 16 - 30
     \]
     Simplify:
     \[
     \left(x^3 + \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8}\right) + \left(xd^2 + \frac{d^3}{2}\right) - \left(x^3 + 3x^2d + 3xd^2 + d^3\right) - \frac{(x + d)d^2}{4} = -14
     \]
     \[
     \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8} + xd^2 + \frac{d^3}{2} - 3x^2d - 3xd^2 - d^3 - \frac{(x + d)d^2}{4} = -14
     \]
     \[
     \frac{3x^2d}{2} + \frac{3xd^2}{4} + \frac{d^3}{8} + xd^2 + \frac{d^3}{2} - 3x^2d - 3xd^2 - d^3 - \frac{(x + d)d^2}{4} = -14
     \]

     Solve for \( x \) and \( d \):
     \[
     x = -\frac{d}{2} \quad \text{or} \quad d = 0
     \]

     If \( d = 0 \):
     \[
     x = y = z = \sqrt[3]{2}
     \]

     If \( x = -\frac{d}{2} \):
     \[
     d = 2\sqrt[3]{2} \implies x = -\sqrt[3]{2}, \quad y = \sqrt[3]{2}, \quad z = 3\sqrt[3]{2}
     \]

The final answer is \( \boxed{ (x, y, z) = (\sqrt[3]{2}, \sqrt[3]{2}, \sqrt[3]{2}) } \) or \( (x, y, z) = (-\sqrt[3]{2}, \sqrt[3]{2}, 3\sqrt[3]{2}) \)