Triples of complex number How many ordered triples of complex 
numbers $(a,b,c)$ are there such that $a^3- 
b$, $b^3-c$, and $c^3-a$ are rational numbers, and 
$a^2(a^4+1)+b^2(b^4+1)+c^2(c^4+1)=2[{(a^3) 
b}+{(b^3)c}+{(c^3)a}]$

$$a^2 (a^4+1)+b^2 (b^4+1)+c^2 (c^4+1)=2[(a^3)b+(b^3)c+(c^3)a] $$ $\implies$ $$(a^6+b^2-2a^3b)+(b^6+c^2-2b^3c)+(c^6+a^2-2c^3a)=0$$ $\implies$ $$(a^3-b)^2+(b^3-c)^2+(c^3-a)^2=0$$ Now as you said these three quantities are rational and as the sum of there squares is $0$ thus they all must equal $0$. Thus the triples are $$(a, a^3, a^9)$$ Can you proceed further from here??
