Let $(a, b, c)$ be a solution of the system. If one of the numbers $a, b, c$ is $0$, e.g. if $c = 0$, then $c(a^2 + b) = a(b + ca)$ leads to $a = 0$, and similarly one gets $b = 0$. Thus $abc = 0$ leads to $a = b = c = 0$ which is indeed a solution. We now look for solutions with $abc \neq 0$. We rewrite the equations
$$\begin{align*} ab(b - c) &= c(c - a) \\ bc(c - a) &= a(a - b) \\ ca(a - b) &= b(b - c). \end{align*}$$
It follows that $a^2b^2c^2(a - b)(b - c)(c - a) = abc(a - b)(b - c)(c - a)$.

Case 1: Among the numbers $a, b, c$ there exist (at least) two equal ones; say $a = b$. We have
$$a^2c + bc = b^2 + abc \Leftrightarrow bc = b^2 \Leftrightarrow c = b,$$
therefore in this case we obtain $a = b = c$. Conversely, any such triple is a solution to the system.

Case 2: If $a \neq b \neq c \neq a$, then $abc = 1$. We obtain $ab(b - c) = c(c - a) \Leftrightarrow (b - c) = c^2(c - a)$ and two more equations, similar to this one. It follows that
$$a^3 + b^3 + c^3 = ac^2 + ba^2 + cb^2.$$
If $a, b, c > 0$, from AM-GM it follows that $a^3 + a^3 + b^3 \ge 3ba^2$ (with equality if and only if $a = b$) which, added with two similar inequalities, leads to $a^3 + b^3 + c^3 \ge ac^2 + ba^2 + cb^2$. We thus have equality in the previous inequality, hence $a = b = c$.

If one of the variables is positive and the other two are negative, say $a > 0$, $b, c < 0$, then
$$b(c^2 + a) = a(a + bc) = a^2 + 1 > 0 \Rightarrow a < 0,$$
contradiction.

In conclusion, the only solutions are $(a, b, c) = (x, x, x)$, $\forall x \in \mathbb{R}$.