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Math Help - a cubic

  1. #1
    Senior Member pacman's Avatar
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    a cubic

    Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
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  2. #2
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    Hello, I got
    a^3+b^3+c^3-3abc = a(a^2-bc)+b(b^2-ac)+c(c^2-ab)=

    =a(a^2-bc -ac -ab +b^2 +c^2)-a(-ac -ab +b^2 +c^2)+
    +b(b^2-ac-bc-ab+a^2+c^2)-b(-bc-ab+a^2+c^2)+
    +c(c^2-ab-ac-bc+a^2+b^2)-c(-ac-bc+a^2+b^2)=

    =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
     +a^2c+a^2b-ab^2-ac^2 +b^2c+ab^2-a^2b-bc^2 +ac^2+bc^2-a^2c-b^2c=

    =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
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  3. #3
    ynj
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    Quote Originally Posted by pacman View Post
    Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
    You only have to remember it:
    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
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  4. #4
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    Quote Originally Posted by ynj View Post
    You only have to remember it:
    a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
    Simply memorizing mathematics without understanding is absurd and idiotic. It is much better to not memorize a certain formula and derive it each time...even if it takes you longer.
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    Quote Originally Posted by RobLikesBrunch View Post
    Simply memorizing mathematics without understanding is absurd and idiotic.
    agree with you, but some things are good to know. I don't remember this formula, but what initially helped me to factor the expression was the formula x^3+y^3 = (x+y)(x^2+y^2 -xy).
    So here is the other way how to factorize, and pacman you'll find another ways if you try.

    a^3+b^3+c^3-3abc = a^3+b^3+c^3-3abc +3ab(a+b) - 3ab(a+b)=
    = (a+b)^3 +c^3 - 3abc -3ab(a+b) = (a+b)^3 +c^3 -3ab(a+b+c) =

    [now we make use of x^3+y^3 = (x+y)(x^2+y^2 -xy)]

    =(a+b+c)((a+b)^2+c^2-(a+b)c) -3ab(a+b+c) =
    =(a+b+c)((a+b)^2+c^2-(a+b)c -3ab) =(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
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  6. #6
    Senior Member pacman's Avatar
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    I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down. Thanks for the efforts to help.
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  7. #7
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    Quote Originally Posted by pacman View Post
    I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down.
    for me, it is about messing around with symbols using identities i already know and recognizing patterns and symmetries. Maybe we should start to learn from computers, i suppose they are better in this than humans. Can anybody explain how computers do it? (i mean how they factorize )
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  8. #8
    Senior Member pacman's Avatar
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    i found this from the web, now i wonder why it is difficult to factor

    The identity would probably be known to Lagrange from his extensive study of algebraic equations If w is a primitive cubic root of unity then a^3 + b^3 + c^3 - 3abc is the constant term of the polynomial satisfied by a +bw +cww
    a^3 + b^3 + c^3 - 3abc = (a+b+c)(a +bw +cww)(a +bww +cw).

    This and other similar identities occur when symmetrical functions of the roots of polynomial
    equations are calculated I know Newton studied symmetric functions of roots. He may have been aware of this identity.


    The (a +bw +cww) is a special form of Eisenstein cubic integrers and a^3 + b^3 + c^3 - 3abc ist its norm. thus, a^3 + b^3 + c^3 - 3abc can also be referrred to a a termary cubic form.
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