$\large (a^2-b^2) $ is proven to be equal to $\large (a+b)(a-b) $ (difference of two squares) . what then is $\large (a^2+b^2) $ and hence expand $\large (a^3+b^3) $ to the same form without the exponents.

P.S. this is my own personal question. I need some understanding

Question

$\large (a^2-b^2) $ is proven to be equal to $\large (a+b)(a-b) $ (difference of two squares) . what then is $\large (a^2+b^2) $ and hence expand $\large (a^3+b^3) $ to the same form without the exponents.

P.S. this is my own personal question. I need some understanding

sasogeek · Answer

by expansion to "the same form without the exponents", i mean, expand it to a form that does not have terms with exponents

zepp · Answer

I guess for the $\large a^2 + b^2$ you can use the complete the square method, let me explain down there.

lalaly · Answer

(a+bi)(a-bi) ?

sasogeek · Answer

where from the i?

lalaly · Answer

do u know complex numbers?

sasogeek · Answer

i've heard of them... not sure if i invited them to this party? how did they get here?

lalaly · Answer

lol thats how i understood ur problem, lol never mind then

lalaly · Answer

because u said terms without exponents

sasogeek · Answer

i said that because difference of two squares reduces to $ (a+b)(a-b) $ and there's no exponents in there... if u see what i mean...

lalaly · Answer

yeah yeah

zepp · Answer

$\large \begin{align} a^2+b^2 \ a^2+b^2 +2ab-2ab \ (a+b)^2-2ab \ \text{Difference of squares} \ (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})\end{align}$
There :)

zepp · Answer

For the cube thingy I don't know xD

sasogeek · Answer

okay thanks for this one :)

zepp · Answer

Wait, I think I have a pdf file somewhere, talking about factorisation, wait a second let me find it.

sasogeek · Answer

does this work?

$\ a^3+b^3 $=(a+b)(a+b)(a+b)

zepp · Answer

I don't think so, I couldn't find it but I think you'll have to expand this into a 2nd degree form with some trashes behind :|

sasogeek · Answer

lol ok i'll see how best i can handle this. i'm sure there's something to it that works :) thanks anyway :)

zepp · Answer

I browsed through the web and found that would be (a+b)(a^2-2ab+b^2), I don't know the steps though :(

and I guess your first step was right, but with a minus sign somewhere lol

sasogeek · Answer

but a^2-2ab+b^2 is the same as (a-b)^2  :/

zepp · Answer

I think I got it :D

zepp · Answer

$\large a^3+b^3$
Since it's a cube, I'll need a^2b and b^2a, I can't really explain why I chose them but :P (it was instinct lol)
$\large a^3+a^2b-a^b+b^2a-b^2a +b^3$
From this, I can factor out all similar stuffs
It would become 
(I'll rearrange that first)
$\large a^3-a^2b+ab^2+a^2b-ab^2+b^3$
There
Now 
$\large a(a^2-ab+b^2)+b(a^2-ab+b^2)$
Thus
$\large (a+b)(a^2-ab+b^2)$

zepp · Answer

I guess I chose $\large a^2b ~~\text{and}~~  b^2a$ because I'll have to reduce a^3 and b^3 to squares, a^2 and b^2
Then since there's 3 a, 3 b, I add b and a, a^2b and b^2a

zepp · Answer

http://puu.sh/uaUs Should be a^2b

across · Answer

zepp is right on.

It's easy to think that$$(a^n+b^n)=(a+b)^n\\exists a,b,n\in\mathbb{Z}.$$This is not true.