$x+x^2+x^3=x^4+x^5+x^6$ 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?

Here is a variant of the solution by @Servaes that seems simpler to me:
Plugging $x=-1$ gives $2=0$ in $A$.
Therefore, $x+x^2+x^3=x^4+x^5+x^6$ is the same as $x+x^2+x^3+x^4+x^5+x^6=0$.
Multiplying that by $x-1$ gives $x^7-x=0$ and so $x^8-x^2=0$.
Now $x^8-x^2=(x^4-x)^2$.
Let $z=x^4-x$. Then $z^2=0$ and $z+z^2+z^3=z^4+z^5+z^6$ imply $z=0$, and so $x^4=x$.
