The equation $x^3 + ax^2 + bx + c = 0$ has three (not necessarily distinct) real roots $t, u, v$. For which $a, b, c$ do the numbers $t^3, u^3, v^3$ satisfy the equation $x^3 + a^3x^2 + b^3x + c^3 = 0$?
1. **Using Viète's Formulas for the First Equation:**
   Given the polynomial \(x^3 + ax^2 + bx + c = 0\) with roots \(t, u, v\), by Viète's formulas, we have:
   \[
   t + u + v = -a,
   \]
   \[
   tu + uv + vt = b,
   \]
   \[
   tuv = -c.
   \]

2. **Using Viète's Formulas for the Second Equation:**
   For the polynomial \(x^3 + a^3x^2 + b^3x + c^3 = 0\) with roots \(t^3, u^3, v^3\), we apply Viète's formulas again:
   \[
   t^3 + u^3 + v^3 = -a^3,
   \]
   \[
   t^3u^3 + u^3v^3 + v^3t^3 = b^3,
   \]
   \[
   t^3u^3v^3 = -c^3.
   \]

3. **Relating the Sums of Cubes:**
   We need to relate the sums of cubes to the original sums. Using the identity for the sum of cubes:
   \[
   t^3 + u^3 + v^3 - 3tuv = (t + u + v)(t^2 + u^2 + v^2 - tu - uv - vt).
   \]
   Since \(t + u + v = -a\), we substitute:
   \[
   t^3 + u^3 + v^3 - 3tuv = -a \left( t^2 + u^2 + v^2 - tu - uv - vt \right).
   \]
   Using the identity \(t^2 + u^2 + v^2 = (t + u + v)^2 - 2(tu + uv + vt)\), we get:
   \[
   t^2 + u^2 + v^2 = a^2 - 2b.
   \]
   Substituting back, we have:
   \[
   t^3 + u^3 + v^3 - 3tuv = -a \left( a^2 - 3b \right).
   \]
   Since \(t^3 + u^3 + v^3 = -a^3\), we equate:
   \[
   -a^3 - 3(-c) = -a(a^2 - 3b).
   \]
   Simplifying, we get:
   \[
   -a^3 + 3c = -a^3 + 3ab.
   \]
   Therefore:
   \[
   3c = 3ab \implies c = ab.
   \]

4. **Relating the Products of Cubes:**
   We need to show that:
   \[
   (tu + uv + vt)^3 = t^3u^3 + u^3v^3 + v^3t^3.
   \]
   Given \(tu + uv + vt = b\), we have:
   \[
   b^3 = t^3u^3 + u^3v^3 + v^3t^3.
   \]
   Since \(tuv = -c\), we have:
   \[
   t^3u^3v^3 = (-c)^3 = -c^3.
   \]
   Therefore, the condition \(c = ab\) ensures that the products of cubes are consistent.

5. **Conclusion:**
   To satisfy both conditions, we must have \(c = ab\). Additionally, for the polynomial \(x^3 + ax^2 + bx + ab = (x+a)(x^2 + b)\) to have real roots, \(b = 0\) is necessary. This gives us the triplet \((a, b, c) = (t, 0, 0)\) where \(t \in \mathbb{R}\).

The final answer is \( \boxed{ (a, b, c) = (t, 0, 0) } \) where \(t \in \mathbb{R}\).