$ 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?