If $ c=0 $ then by $ c(a^2+b)=b(b+ca) $ we get $ b=0 $ and then it's trivial that $ a=0 $. Thus, we consider the case that there isn't a zero, now. Notice the following equalities are equivalent to the original conditions. \[ ab(b-c)=c(c-a) \] \[ bc(c-a)=a(a-b) \] \[ ca(a-b)=b(b-c) \] So, we have $ a^2b^2c^2(a-b)(b-c)(c-a)=abc(a-b)(b-c)(c-a) $ Case 1: There are two equal numbers $ a,b $ Then from the condition $ (3) $, we have \[ a^2c+bc=b^2+abc \Leftrightarrow bc=b^2 \Leftrightarrow c=b \] Case 2: $ abc=1 $ Then $ ab(b-c)=c(c-a) \Leftrightarrow (b-c)=c^2(c-a) $, and the others are similar. \[ \Rightarrow a^3+b^3+c^3=ac^2+ba^2+cb^2 \] If $ a,b,c>0 $ then $ a=b=c $ by using Muirhead inequality. If $ a>0>b>c $ then... \[ b(c^2+a)=a(a+bc)=a^2+1>0 \Rightarrow a<0 \] , which is impossible! In conclusion, the solutions are $ (a,b,c)=(x,x,x) \forall x \in \mathbb{R} $

How do you apply Muirhead here? the rhs polynomial isn't complete in case 2 $(abc = 1)$. Or am I missing something?

Maybe (s)he meant AM-GM instead . Use $$ac^2 \leq \frac{a^3+c^3+c^3}{3}$$ and the analogues and then add them

$ a^2b^2c^2(a-b)(b-c)(c-a)=abc(a-b)(b-c)(c-a) \iff (abc-1)abc(a-b)(b-c)(c-a)=0$ If $0\in\lbrace a,b,c\rbrace$ then $0=a=b=c$ Assume now that none of $a,b,c$ is zero. If $a=b\vee b=c\vee c=a$ then by system of three equations as in post 2 we get $a=b=c$. Now assume that none of $a,b,c$ is zero and every two of them are different. Then $abc=1\implies b-c=c^2(c-a)\wedge c-a=a^2(a-b)\wedge a-b=b^2(b-c)$ Take WLOG $a>b\wedge a>c$. Then in second equation we have $c-a<0\wedge a^2(a-b)>0$ contradiction. After checking the only solution is $a=b=c$.

ComplexPhi wrote: Maybe (s)he meant AM-GM instead . Use $$ac^2 \leq \frac{a^3+c^3+c^3}{3}$$and the analogues and then add them am gm for reals?????

And here is the solution set over complex numbers: $(t,t,t)$ with $t\in\mathbb{C}$ and cycles of $(A_1,A_2,A_3)$, $(\bar{A}_1,\bar{A}_2,\bar{A}_3)$ $A_n=\frac{1+i}{3}-\frac{e^{2\pi ni/3}}{6}\sqrt[3]{-56+36\sqrt{3}+\left(20-24\sqrt{3}\right)i}-\frac{e^{2\pi ni/3}}{6}\sqrt[3]{56+36\sqrt{3}-\left(20+24\sqrt{3}\right)i}$