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sasogeek
 3 years 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
sasogeek
 3 years 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

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

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

lalaly
 3 years ago
Best ResponseYou've already chosen the best response.0do u know complex numbers?

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

lalaly
 3 years ago
Best ResponseYou've already chosen the best response.0lol thats how i understood ur problem, lol never mind then

lalaly
 3 years ago
Best ResponseYou've already chosen the best response.0because u said terms without exponents

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0i 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...

zepp
 3 years ago
Best 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 :)

zepp
 3 years ago
Best ResponseYou've already chosen the best response.3For the cube thingy I don't know xD

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0okay thanks for this one :)

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

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

zepp
 3 years ago
Best ResponseYou've already chosen the best response.3I 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
 3 years ago
Best ResponseYou've already chosen the best response.0lol ok i'll see how best i can handle this. i'm sure there's something to it that works :) thanks anyway :)

zepp
 3 years ago
Best ResponseYou've already chosen the best response.3I 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

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0but a^22ab+b^2 is the same as (ab)^2 :/

zepp
 3 years ago
Best 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 \(\large (a+b)(a^2ab+b^2)\)

zepp
 3 years ago
Best ResponseYou've already chosen the best response.3I 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
 3 years ago
Best ResponseYou've already chosen the best response.3http://puu.sh/uaUs Should be a^2b

across
 2 years ago
Best ResponseYou've already chosen the best response.0zepp 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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