# \$ x + x ^ 2 + x ^ 3 = x ^ 4 + x ^ 5 + x ^ 6 \$ implies \$ x ^ 4 = x \$ in a ring

Let $$(A, +,.)$$ to be a ring $$x + x ^ 2 + x ^ 3 = x ^ 4 + x ^ 5 + x ^ 6$$ for everyone $$x in A$$. Prove it $$x ^ 4 = x$$ for any x in A. Can any one give me any advice, please?