- #1

donglepuss

- 15

- 1

- TL;DR Summary
- ##(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?

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

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- Thread starter donglepuss
- Start date

- #1

donglepuss

- 15

- 1

- TL;DR Summary
- ##(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?

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

Last edited by a moderator:

- #2

- 17,847

- 19,121

Yes. Even for rational or real numbers.

- #3

Mark44

Mentor

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- 8,967

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.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##

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.

- #4

- 17,847

- 19,121

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^3(b^2)-ya^3+xb^2-xy##

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

- #5

Mark44

Mentor

- 36,911

- 8,967

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

- #6

WWGD

Science Advisor

Gold Member

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