Does This Equation Hold in All Commutative Rings?

[donglepuss]

TL;DR

$(a^3+x)(b^2-y)=a^3(b^2)-ya^3+xb^2-xy$

$(a^3+x)(b^2-y)=a^3(b^2)-ya^3+xb^2-xy$

is this correct for all whole numbers x,y,a,b?

 

[fresh_42]

Yes. Even for rational or real numbers.

 

[Mark44]

is this correct for all whole numbers x,y,a,b?
$(a^3+x)(b^2-y)=a^3(b^2)-ya^3+xb^2-xy$

This equation is a special kind of equation: an identity, one that is true for all values of the variables, whether whole numbers, integers, rationals, reals, or complex numbers. It's even true for square matrices, as long as they are all the same size.

A simpler example of an identity is this: $a(b + c) = ab + ac$, which is true for any mathematical structures that support multiplication distributing over addition.

 

[fresh_42]

Your equation isn't written in its best form. If you compare @Mark44's formula with what you have written, then

$(a^3+x)(b^2-y)=a^3(b^2)-ya^3+xb^2-xy$

is a bit inaccurate. There are structures in mathematics which are in general not commutative, rings and algebras. So it is a good habit to learn it by respecting left and right, so it's better to write
$(a^3+x)(b^2-y)=a^3b^2-a^3y+xb^2-xy$

 

[Mark44]

is a bit inaccurate.

I didn't notice that $a^3$ times $-y$ was written as $-ya^3$, when I mentioned matrix multiplication, which isn't generally commutative.

 

[WWGD]

Notice how the general binomial $(a+b)^n = \Sigma _{i=0}^n (nCi )x^n y^i$ ; $nCi$ := " n choose i" also assumes commutativity. I think the OP and this identity hold for all commutative rings.