just a sketch of proof. Let $ y_1(x) = ax^3+bx+c$, $ y_2(x) = bx^3+cx+a$, $ y_3(x) = cx^3+ax+b$, it results that $(1, a+b+c)$ is a common point of $ y_1, y_2$ and $y_3$. One can prove that this is the only common point. If $\alpha$ is s.t. $y_1(\alpha)=y_2(\alpha)=y_3(\alpha)$, we have simultaneously that is $y_1(\alpha)-y_2(\alpha)=0$, $y_2(\alpha)-y_3(\alpha)=0$. From which $(b-c)(y_1(\alpha)-y_2(\alpha)) - (a-b)(y_2(\alpha)-y_3(\alpha))=0$ Then the common root must be $x = 1$, from which it follows that $a+b+c = 0$. Rewriting then the equations: $ ax^3+bx-(a+b)=0$, $ bx^3-(a+b)x+a=0$, $ -(a+b)x^3+ax+b=0$ which, for $x \neq1$, are equivalent to: $ ax^2+ax + a+b=0$, $bx^2+bx-a=0$, and $(a+b)x^2+(a+b)x+b=0$. Now it is sufficient to prove that at least one of the following discriminant $D_i$ is not negative. $D_1= -a(3a+4b)$, $D_2= b(4a+b)$ and $D_3= (a+b)(a-3b)$.