A special equation in a Boolean ring

Let us consider a ring R such that ∀ x ∈ R x²=x (such a ring is called Boolean). Let us solve the equation in R x R x R:

(XY)⁶-(YX)⁶=2X⁷-4Y⁵+8YZ³

Let us just recall that a ring R is a set provided with two binary operations + and • such that (R,+) is an abelian group, (R,•) is a semigroup and • is distributive with respect to the sum. Let us prove that the set of solutions of this equation is R x R x R. Let us observe that if a∈R we have that 2a=0. In fact:

2a=a+a=(a+a)²=(a+a)(a+a)=a²+a²+a²+a²=4a=2a+2a da cui 2a=0.

Now let’s note that R is commutative. If a and b are in R we have that:

a+b=(a+b)²=a²+ab+ba+b²=a+b+ab+ba so ab+ba=0 and so ab+ba=ba+ba which means ab=ba.

Let now (X,Y,Z)∈R x R x R,then:

(XY)⁶-(YX)⁶ = (XY)⁶-(XY)⁶ = 0

2X⁷-4Y⁵+8YZ³ = 0+0+0=0

It follows that (X,Y,Z) is a solution . Then :

{(X,Y,Z) ∈ R x R x R : (XY)⁶-(YX)⁶=2X⁷-4Y⁵+8YZ³ } = R³