Find all real triples $(a,b,c)$, for which $$a(b^2+c)=c(c+ab)$$ $$b(c^2+a)=a(a+bc)$$ $$c(a^2+b)=b(b+ca).$$
We are given the system of equations:
\[ a(b^2 + c) = c(c + ab) \]
\[ b(c^2 + a) = a(a + bc) \]
\[ c(a^2 + b) = b(b + ca) \]

We need to find all real triples \((a, b, c)\) that satisfy these equations.

1. **Case 1: One of the variables is zero**

   Suppose \(c = 0\). Then the third equation becomes:
   \[ c(a^2 + b) = b(b + ca) \]
   \[ 0 = b^2 \]
   This implies \(b = 0\). Substituting \(b = 0\) and \(c = 0\) into the first equation:
   \[ a(0 + 0) = 0(0 + a \cdot 0) \]
   \[ 0 = 0 \]
   This is trivially true for any \(a\). Therefore, one solution is \((a, b, c) = (a, 0, 0)\) for any real number \(a\).

2. **Case 2: None of the variables are zero**

   We rewrite the given equations in a more symmetric form:
   \[ a(b^2 + c) = c(c + ab) \]
   \[ b(c^2 + a) = a(a + bc) \]
   \[ c(a^2 + b) = b(b + ca) \]

   By manipulating these equations, we can derive:
   \[ ab(b - c) = c(c - a) \]
   \[ bc(c - a) = a(a - b) \]
   \[ ca(a - b) = b(b - c) \]

   Multiplying these three equations together, we get:
   \[ (ab(b - c))(bc(c - a))(ca(a - b)) = (c(c - a))(a(a - b))(b(b - c)) \]
   \[ a^2b^2c^2(b - c)(c - a)(a - b) = abc(a - b)(b - c)(c - a) \]

   Dividing both sides by \(abc(a - b)(b - c)(c - a)\) (assuming \(a \neq b\), \(b \neq c\), and \(c \neq a\)), we get:
   \[ abc = 1 \]

3. **Case 3: \(abc = 1\)**

   Substituting \(abc = 1\) into the original equations, we get:
   \[ ab(b - c) = c(c - a) \]
   \[ bc(c - a) = a(a - b) \]
   \[ ca(a - b) = b(b - c) \]

   Adding these three equations, we get:
   \[ ab(b - c) + bc(c - a) + ca(a - b) = c(c - a) + a(a - b) + b(b - c) \]
   \[ ab(b - c) + bc(c - a) + ca(a - b) = ab(b - c) + bc(c - a) + ca(a - b) \]

   This is an identity, which means the equations are consistent under the condition \(abc = 1\).

4. **Conclusion**

   From the above cases, we conclude that the solutions are:
   \[ (a, b, c) = (x, x, x) \text{ for any } x \in \mathbb{R} \]
   and
   \[ (a, b, c) = (a, 0, 0) \text{ for any } a \in \mathbb{R} \]

The final answer is \( \boxed{ (a, b, c) = (x, x, x) \text{ for any } x \in \mathbb{R} } \) and \((a, b, c) = (a, 0, 0) \text{ for any } a \in \mathbb{R}\).