Question

why ((a+b)^2)*((a-b)^2) is equal (a^2 -b^2)^2?

Answer 1

Hey Frederico :)
When we multiply `conjugates`,\[\Large\rm (x-y)(x+y)=x^2\cancel{-xy}\cancel{+xy}-y^2\]\[\Large\rm =x^2-y^2\]We always end up with the `difference of squares`.

So notice that in your problem, you're actually starting with:\[\Large\rm (a+b)(a-b)\cdot (a+b)(a-b)\]Yes? :)

Answer 2

I guess is (a+b)(a+b)*(a+b)(a-b). Is not it?

Answer 3

Yes, but recall that multiplication is `commutative`, we can multiply in any order.
So I was changing the order so it would match this idea of "conjugates"

Answer 4

\[\Large\rm \color{orangered}{(a-b)(a+b)}\cdot (a-b)(a+b)\]So what you wrote is the same as what is written here :)
So if we use our rule that was posted above we get:\[\Large\rm \color{orangered}{(a^2-b^2)}\cdot (a-b)(a+b)\]Yes?
Or still stuck on that weird ordering of the multiplication?

Answer 5

patience please... hahhahah

Answer 6

in my equations I hav 3 plus and 1 minus. in yours you rave two minus and two plus. I didnt get it.

Answer 7

No no, in yours you have 2 plus, 2 minus,\[\Large\rm (a+b)^2=(a+b)(a+b)\]\[\Large\rm (a-b)^2=(a-b)(a-b)\]In your original question I mean*

Answer 8

yessssssssssssss! now I get it! Sorry and thank you!!!

Answer 9

ah cool :)