Proof that ring is commutative

Question

Proof that a ring $(A, + , \cdot)$ is commutative if $$x(y^2+y)=(y^2+y)x$$

I set $x$ to be $$(x-xy^2) $$and I get that $$xy^3=yxy^2.$$But I don't know how to get $xy=yx$.

Answer 1

Let $Z(A) = \{x \in A| yx = xy, \forall y \in A \}$.

The hypothesis means that $x^2 + x \in Z(A), \forall x \in A$.

Now set $x := x+y$ and we get that $$(x+y)^2 + (x+y) \in Z(A) \iff x^2 + xy + yx + y^2 + x + y \in Z(A) \iff $$

$$ \iff (x^2 + x) + (y^2 + y) + (xy+yx) \in Z(A) \implies xy + yx \in Z(A).$$

This means that $$\forall x,y \in A, x(xy+yx) = (xy+yx)x \iff x^2y + xyx = xyx + yx^2 \iff x^2y = yx^2.$$

The hypothesis is rewritten as $$xy^2 + xy = y^2x + yx, \forall x, y \in A,$$ and using the relation from above, we get that $$xy = yx, \forall x,y \in A,$$ so $A$ is a commutative ring.