We are given the system of equations:
$$\begin{cases} x^3 = 3x - 12y + 50, \\ y^3 = 12y + 3z - 2, \\ z^3 = 27z + 27x. \end{cases}$$

To find all triples $(x, y, z)$ of real numbers that satisfy these equations, we analyze the behavior of the functions involved.

1. **Case $x > 2$:**
- For $x > 2$, the function $x^3 - 3x$ is increasing. Thus, from the first equation, $y < 4$.
- From the third equation, since $x > 2$, we have $z^3 - 27z > 54$, implying $z > 6$.
- However, substituting $z > 6$ into the second equation gives $y^3 - 12y > 16$, implying $y > 4$, which contradicts $y < 4$.

2. **Case $x < 2$:**
- For $x < 2$, from the third equation, $z^3 - 27z < 54$, implying $z < 6$.
- From the second equation, $y < 4$.
- Substituting $y < 4$ into the first equation gives $x^3 - 3x = 50 - 12y > 2$, implying $x > 2$, which contradicts $x < 2$.

3. **Verification of $x = 2, y = 4, z = 6$:**
- Substituting $x = 2$, $y = 4$, and $z = 6$ into the equations:
$$2^3 = 3(2) - 12(4) + 50 \implies 8 = 6 - 48 + 50 \implies 8 = 8,$$
$$4^3 = 12(4) + 3(6) - 2 \implies 64 = 48 + 18 - 2 \implies 64 = 64,$$
$$6^3 = 27(6) + 27(2) \implies 216 = 162 + 54 \implies 216 = 216.$$
- These values satisfy all three equations.

Thus, the only solution is:
$$(x, y, z) = \boxed{(2, 4, 6)}.$$