Find all natural triples $(a,b,c)$, such that:
$a - )\,a \le b \le c$
$b - )\,(a,b,c) = 1$
$c - )\,\left. {{a^2}b} \right|{a^3} + {b^3} + {c^3}\,,\,\left. {{b^2}c} \right|{a^3} + {b^3} + {c^3}\,,\,\left. {{c^2}a} \right|{a^3} + {b^3} + {c^3}$.
1. **Given Conditions:**
   - We need to find all natural triples \((a, b, c)\) such that:
     \[
     a - )\,a \le b \le c
     \]
     \[
     b - )\,(a, b, c) = 1
     \]
     \[
     c - )\,\left. {{a^2}b} \right|{a^3} + {b^3} + {c^3}\,,\,\left. {{b^2}c} \right|{a^3} + {b^3} + {c^3}\,,\,\left. {{c^2}a} \right|{a^3} + {b^3} + {c^3}
     \]

2. **Analyzing the GCD Condition:**
   - Let \((a, b) = d\). Since \({a^2}b \left| {a^3} + {b^3} + {c^3}\), it implies \(d \left| c\).
   - Given \((a, b, c) = 1\), it follows that \(d = 1\). Hence, \(a, b, c\) are pairwise coprime.

3. **Divisibility Condition:**
   - Since \(a, b, c\) are pairwise coprime, we have:
     \[
     {a^2}{b^2}{c^2} \left| {a^3} + {b^3} + {c^3}
     \]
   - This implies:
     \[
     {a^2}{b^2}{c^2} \leq {a^3} + {b^3} + {c^3} \leq 3{c^3}
     \]
   - Therefore:
     \[
     c \geq \frac{{a^2}{b^2}}{3}
     \]

4. **Further Analysis:**
   - Given \({c^2} \left| {a^3} + {b^3}\), we have:
     \[
     2{b^3} \geq {a^3} + {b^3} \geq {c^2} \geq \left(\frac{{a^2}{b^2}}{3}\right)^2
     \]
   - This simplifies to:
     \[
     18 \geq {a^4}b \geq {a^5}
     \]
   - Hence, \(a = 1\).

5. **Case Analysis for \(b\):**
   - If \(b = 1\):
     - From \({c^2} \left| {a^3} + {b^3} = 2\), we have \(c = 1\).
   - If \(b > 1\):
     - Since \((b, c) = 1\), it follows that \(c \geq b + 1\).
     - Therefore:
       \[
       2{c^3} > {c^3} + {b^3} + 1 = {a^3} + {b^3} + {c^3} \geq {b^2}{c^2}
       \]
     - This implies:
       \[
       c > \frac{{b^2}}{2}
       \]
     - Additionally:
       \[
       1 + {b^3} = {a^3} + {b^3} \geq {c^2} \geq \left(\frac{{b^2}}{2}\right)^2
       \]
     - Hence:
       \[
       b \leq 4
       \]

6. **Finding \(b\) and \(c\):**
   - From \({c^2} \left| {a^3} + {b^3} = {b^3} + 1\), we find:
     - For \(b = 2\), \(c = 3\).

7. **Conclusion:**
   - The solutions are \((a, b, c) = (1, 1, 1)\) and \((1, 2, 3)\).

The final answer is \( \boxed{ (1, 1, 1) } \) and \((1, 2, 3)\).