Determine all triples of real numbers $(a,b,c)$ that satisfy simultaneously the equations:
$$\begin{cases}  a(b^2 + c) = c(c + ab) \\ b(c^2 + a) = a(a + bc) \\ c(a^2 + b) = b(b + ca) \end{cases}$$
To determine all triples of real numbers \((a, b, c)\) that satisfy the given system of equations:
\[
\begin{cases}
a(b^2 + c) = c(c + ab) \\
b(c^2 + a) = a(a + bc) \\
c(a^2 + b) = b(b + ca)
\end{cases}
\]

1. **Simplify the first equation:**
   \[
   a(b^2 + c) = c(c + ab)
   \]
   Expanding both sides, we get:
   \[
   ab^2 + ac = c^2 + abc
   \]
   Rearranging terms, we have:
   \[
   ab^2 + ac - abc - c^2 = 0
   \]
   Factoring out common terms:
   \[
   ab(b - c) = c(c - a)
   \]

2. **Simplify the second equation:**
   \[
   b(c^2 + a) = a(a + bc)
   \]
   Expanding both sides, we get:
   \[
   bc^2 + ba = a^2 + abc
   \]
   Rearranging terms, we have:
   \[
   bc^2 + ba - a^2 - abc = 0
   \]
   Factoring out common terms:
   \[
   bc(c - a) = a(a - b)
   \]

3. **Simplify the third equation:**
   \[
   c(a^2 + b) = b(b + ca)
   \]
   Expanding both sides, we get:
   \[
   ca^2 + cb = b^2 + bca
   \]
   Rearranging terms, we have:
   \[
   ca^2 + cb - b^2 - bca = 0
   \]
   Factoring out common terms:
   \[
   ca(a - b) = b(b - c)
   \]

4. **Analyze the simplified equations:**
   We now have the following system of equations:
   \[
   \begin{cases}
   ab(b - c) = c(c - a) \\
   bc(c - a) = a(a - b) \\
   ca(a - b) = b(b - c)
   \end{cases}
   \]

5. **Consider the case \(a = b = c\):**
   If \(a = b = c\), then substituting into any of the original equations, we get:
   \[
   a(a^2 + a) = a(a + a^2)
   \]
   which simplifies to:
   \[
   a^3 + a^2 = a^3 + a^2
   \]
   This is always true, so \(a = b = c\) is a solution.

6. **Consider the case where \(a, b, c\) are distinct:**
   Suppose \(a, b, c\) are distinct. From the simplified equations, we have:
   \[
   ab(b - c) = c(c - a)
   \]
   If \(a = 0\), then:
   \[
   0 = c^2
   \]
   which implies \(c = 0\). Similarly, \(b = 0\). Thus, \(a = b = c = 0\).

   If \(a, b, c \neq 0\), we can divide the equations by \(abc\) to get:
   \[
   \frac{b(b - c)}{c} = \frac{c - a}{a}
   \]
   \[
   \frac{c(c - a)}{a} = \frac{a - b}{b}
   \]
   \[
   \frac{a(a - b)}{b} = \frac{b - c}{c}
   \]
   This implies:
   \[
   \frac{b^2 - bc}{c} = \frac{c - a}{a}
   \]
   \[
   \frac{c^2 - ca}{a} = \frac{a - b}{b}
   \]
   \[
   \frac{a^2 - ab}{b} = \frac{b - c}{c}
   \]
   These equations are symmetric and suggest that \(a = b = c\) is the only solution.

Conclusion:
The only solution to the system of equations is \(a = b = c\).

The final answer is \( \boxed{ a = b = c } \).