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\(\large (a^2b^2) \) is proven to be equal to \(\large (a+b)(ab) \) (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
 one year ago
 one year ago
\(\large (a^2b^2) \) is proven to be equal to \(\large (a+b)(ab) \) (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
 one year ago
 one year ago

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sasogeekBest ResponseYou've already chosen the best response.0
by expansion to "the same form without the exponents", i mean, expand it to a form that does not have terms with exponents
 one year ago

zeppBest ResponseYou've already chosen the best response.3
I guess for the \(\large a^2 + b^2\) you can use the complete the square method, let me explain down there.
 one year ago

lalalyBest ResponseYou've already chosen the best response.0
do u know complex numbers?
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
i've heard of them... not sure if i invited them to this party? how did they get here?
 one year ago

lalalyBest ResponseYou've already chosen the best response.0
lol thats how i understood ur problem, lol never mind then
 one year ago

lalalyBest ResponseYou've already chosen the best response.0
because u said terms without exponents
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
i said that because difference of two squares reduces to \( (a+b)(ab) \) and there's no exponents in there... if u see what i mean...
 one year ago

zeppBest ResponseYou've already chosen the best response.3
\(\large \begin{align} a^2+b^2 \\ a^2+b^2 +2ab2ab \\ (a+b)^22ab \\ \text{Difference of squares} \\ (a+b+\sqrt{2ab})(a+b\sqrt{2ab})\end{align}\) There :)
 one year ago

zeppBest ResponseYou've already chosen the best response.3
For the cube thingy I don't know xD
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
okay thanks for this one :)
 one year ago

zeppBest ResponseYou've already chosen the best response.3
Wait, I think I have a pdf file somewhere, talking about factorisation, wait a second let me find it.
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
does this work? \(\ a^3+b^3 \)=(a+b)(a+b)(a+b)
 one year ago

zeppBest ResponseYou've already chosen the best response.3
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 :
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
lol ok i'll see how best i can handle this. i'm sure there's something to it that works :) thanks anyway :)
 one year ago

zeppBest ResponseYou've already chosen the best response.3
I browsed through the web and found that would be (a+b)(a^22ab+b^2), I don't know the steps though :( and I guess your first step was right, but with a minus sign somewhere lol
 one year ago

sasogeekBest ResponseYou've already chosen the best response.0
but a^22ab+b^2 is the same as (ab)^2 :/
 one year ago

zeppBest ResponseYou've already chosen the best response.3
\(\large a^3+b^3\) 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) \(\large a^3+a^2ba^b+b^2ab^2a +b^3\) From this, I can factor out all similar stuffs It would become (I'll rearrange that first) \(\large a^3a^2b+ab^2+a^2bab^2+b^3\) There Now \(\large a(a^2ab+b^2)+b(a^2ab+b^2)\) Thus \(\large (a+b)(a^2ab+b^2)\)
 one year ago

zeppBest ResponseYou've already chosen the best response.3
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
 one year ago

zeppBest ResponseYou've already chosen the best response.3
http://puu.sh/uaUs Should be a^2b
 one year ago

acrossBest ResponseYou've already chosen the best response.0
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.
 one year ago
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