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sasogeek

  • 2 years ago

\(\large (a^2-b^2) \) 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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  1. sasogeek
    • 2 years ago
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    by expansion to "the same form without the exponents", i mean, expand it to a form that does not have terms with exponents

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

  3. lalaly
    • 2 years ago
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    (a+bi)(a-bi) ?

  4. sasogeek
    • 2 years ago
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    where from the i?

  5. lalaly
    • 2 years ago
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    do u know complex numbers?

  6. sasogeek
    • 2 years ago
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    i've heard of them... not sure if i invited them to this party? how did they get here?

  7. lalaly
    • 2 years ago
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    lol thats how i understood ur problem, lol never mind then

  8. lalaly
    • 2 years ago
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    because u said terms without exponents

  9. sasogeek
    • 2 years ago
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    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...

  10. lalaly
    • 2 years ago
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    yeah yeah

  11. zepp
    • 2 years ago
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    \(\large \begin{align} a^2+b^2 \\ a^2+b^2 +2ab-2ab \\ (a+b)^2-2ab \\ \text{Difference of squares} \\ (a+b+\sqrt{2ab})(a+b-\sqrt{2ab})\end{align}\) There :)

  12. zepp
    • 2 years ago
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    For the cube thingy I don't know xD

  13. sasogeek
    • 2 years ago
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    okay thanks for this one :)

  14. zepp
    • 2 years ago
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    Wait, I think I have a pdf file somewhere, talking about factorisation, wait a second let me find it.

  15. sasogeek
    • 2 years ago
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    does this work? \(\ a^3+b^3 \)=(a+b)(a+b)(a+b)

  16. zepp
    • 2 years ago
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    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 :|

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

  18. zepp
    • 2 years ago
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    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

  19. sasogeek
    • 2 years ago
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    but a^2-2ab+b^2 is the same as (a-b)^2 :/

  20. zepp
    • 2 years ago
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    I think I got it :D

  21. zepp
    • 2 years ago
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    \(\large a^3+b^3\) 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 \(\large (a+b)(a^2-ab+b^2)\)

  22. zepp
    • 2 years ago
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    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

  23. zepp
    • 2 years ago
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    http://puu.sh/uaUs Should be a^2b

  24. across
    • 2 years ago
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    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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