If $(A, +, •)$ is a ring and $(x^2+x+1)y = y(x^2+x+1)$, $\forall x, y \in A$ then $A$ is commutative.
Book answer: We put $x \to - x$ and we have $2(xy-yx)=0$ then $xy=yx$. How do I know $2$ is not divisor of $0$?
If $(A, +, •)$ a ring with unity such that $x^3=x^2, \forall x \in A$. Show that $x^2 = x$ and $A$ is commutative.
Book answer: We put $x \to 1$ and we have $1=-1$ and $x+x=0$. After we put $x \to x+1$ we have $3x^2 + 3x = 0$ then $x^2 = x$. How do I know $3$ is not divisor of $0$?