<s>[INST] $x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$  for all $x  \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?
 [/INST] Hint 1: Plugging in $x=-1$ shows that $2=0$ in $A$. 
Hint 2: The given equation can be rearranged to get

 $$(x+x^2+x^3)(1-x^3)=0,$$
 or equivalently $(x^4-x)(1+x+x^2)=0$. 
 Multiplying through by $x-x^2$ shows that
 $$(x^4-x)^2=0.$$

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