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- sasogeek

\(\large (a^2-b^2) \) 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

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- sasogeek

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- sasogeek

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

- zepp

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

- lalaly

(a+bi)(a-bi) ?

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- sasogeek

where from the i?

- lalaly

do u know complex numbers?

- sasogeek

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

- lalaly

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

- lalaly

because u said terms without exponents

- sasogeek

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

yeah yeah

- zepp

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

- zepp

For the cube thingy I don't know xD

- sasogeek

okay thanks for this one :)

- zepp

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

- sasogeek

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

- zepp

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

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

- zepp

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

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

- zepp

I think I got it :D

- zepp

\(\large a^3+b^3\)
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

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

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

- across

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.

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