Find all triples of real numbers $(a, b, c)$ satisfying $$a+b+c=14, \quad a^2+b^2+c^2=84,\quad a^3+b^3+c^3=584.$$
Given the equations:
\[ a + b + c = 14, \]
\[ a^2 + b^2 + c^2 = 84, \]
\[ a^3 + b^3 + c^3 = 584. \]

1. **Calculate \(ab + bc + ca\):**
   \[
   (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
   \]
   Substituting the given values:
   \[
   14^2 = 84 + 2(ab + bc + ca)
   \]
   \[
   196 = 84 + 2(ab + bc + ca)
   \]
   \[
   196 - 84 = 2(ab + bc + ca)
   \]
   \[
   112 = 2(ab + bc + ca)
   \]
   \[
   ab + bc + ca = 56
   \]

2. **Calculate \((a+b)(b+c)(c+a)\):**
   \[
   (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)
   \]
   Substituting the given values:
   \[
   14^3 = 584 + 3(a + b)(b + c)(c + a)
   \]
   \[
   2744 = 584 + 3(a + b)(b + c)(c + a)
   \]
   \[
   2744 - 584 = 3(a + b)(b + c)(c + a)
   \]
   \[
   2160 = 3(a + b)(b + c)(c + a)
   \]
   \[
   (a + b)(b + c)(c + a) = 720
   \]

3. **Form the polynomial whose roots are \(a+b, b+c, c+a\):**
   Let \(x = a+b, y = b+c, z = c+a\). Then:
   \[
   x + y + z = 2(a + b + c) = 2 \times 14 = 28
   \]
   \[
   xy + yz + zx = (a+b)(b+c) + (b+c)(c+a) + (c+a)(a+b) = a^2 + b^2 + c^2 + ab + bc + ca = 84 + 56 = 140
   \]
   \[
   xyz = (a+b)(b+c)(c+a) = 720
   \]

   The polynomial with roots \(x, y, z\) is:
   \[
   t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0
   \]
   Substituting the values:
   \[
   t^3 - 28t^2 + 252t - 720 = 0
   \]

4. **Solve the polynomial:**
   By inspection or using the Rational Root Theorem, we find that the roots are \(6, 10, 12\). Thus, \(a+b, b+c, c+a\) can be \(6, 10, 12\) in some order.

5. **Determine \(a, b, c\) from the roots:**
   We have three cases to consider:
   - \(a+b = 6, b+c = 10, c+a = 12\)
   - \(a+b = 6, b+c = 12, c+a = 10\)
   - \(a+b = 10, b+c = 6, c+a = 12\)

   Solving each case:

   **Case 1:**
   \[
   a + b = 6 \quad (1)
   \]
   \[
   b + c = 10 \quad (2)
   \]
   \[
   c + a = 12 \quad (3)
   \]

   Adding equations (1) and (2):
   \[
   a + 2b + c = 16
   \]
   Subtracting equation (3):
   \[
   a + 2b + c - (c + a) = 16 - 12
   \]
   \[
   2b = 4
   \]
   \[
   b = 2
   \]

   Substituting \(b = 2\) into equations (1) and (2):
   \[
   a + 2 = 6 \implies a = 4
   \]
   \[
   2 + c = 10 \implies c = 8
   \]

   Thus, one solution is \((a, b, c) = (4, 2, 8)\).

   **Case 2:**
   \[
   a + b = 6 \quad (1)
   \]
   \[
   b + c = 12 \quad (2)
   \]
   \[
   c + a = 10 \quad (3)
   \]

   Adding equations (1) and (2):
   \[
   a + 2b + c = 18
   \]
   Subtracting equation (3):
   \[
   a + 2b + c - (c + a) = 18 - 10
   \]
   \[
   2b = 8
   \]
   \[
   b = 4
   \]

   Substituting \(b = 4\) into equations (1) and (2):
   \[
   a + 4 = 6 \implies a = 2
   \]
   \[
   4 + c = 12 \implies c = 8
   \]

   Thus, another solution is \((a, b, c) = (2, 4, 8)\).

   **Case 3:**
   \[
   a + b = 10 \quad (1)
   \]
   \[
   b + c = 6 \quad (2)
   \]
   \[
   c + a = 12 \quad (3)
   \]

   Adding equations (1) and (2):
   \[
   a + 2b + c = 16
   \]
   Subtracting equation (3):
   \[
   a + 2b + c - (c + a) = 16 - 12
   \]
   \[
   2b = 4
   \]
   \[
   b = 2
   \]

   Substituting \(b = 2\) into equations (1) and (2):
   \[
   a + 2 = 10 \implies a = 8
   \]
   \[
   2 + c = 6 \implies c = 4
   \]

   Thus, the final solution is \((a, b, c) = (8, 2, 4)\).

The final answer is \( \boxed{ (4, 2, 8), (2, 4, 8), (8, 2, 4) } \).