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Thread: Factoring

  1. #1
    Junior Member DIOGYK's Avatar
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    Factoring

    I have another factoring problem from Gelfand's Algebra book that I've been unable to solve;

    Problem 122. Factor:
    (f) $\displaystyle (a-b)^3+(b-c)^3+(c-a)^3$

    My try:
    $\displaystyle (a-b)^3+(b-c)^3+(c-a)^3=((a-b)+(b-c))((a-b)^2-(a-b)(b-c)+(b-c)^2)+(c-a)^3=(a-c)(a^2-2ab+b^2-(ab-ac-b^2+bc)+b^2-2bc-c^2)+(c-a)^3=(a-c)(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2)-(a-c)^3=(a-c)(a^2-3ab+3b^2+ac-3bc+c^2)-(a-c)^3=?$

    Thanks in advance.
    Last edited by DIOGYK; Aug 27th 2012 at 10:06 AM.
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  2. #2
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    Re: Factoring

    One of your previous problems,
    Another factoring problem

    shows that $\displaystyle x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx)$

    so if $\displaystyle x+y+z = 0$, then $\displaystyle x^3 + y^3 + z^3 = 3xyz$.

    Maybe you can apply this result to the current problem.
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  3. #3
    Junior Member DIOGYK's Avatar
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    Re: Factoring

    Quote Originally Posted by awkward View Post
    One of your previous problems,
    Another factoring problem

    shows that $\displaystyle x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx)$

    so if $\displaystyle x+y+z = 0$, then $\displaystyle x^3 + y^3 + z^3 = 3xyz$.

    Maybe you can apply this result to the current problem.
    So I guess the answer is $\displaystyle 3(a-b)(b-c)(c-a)$?

    But isn't $\displaystyle x^3+y^3+z^3$ equal to $\displaystyle 3xyz$ only if $\displaystyle x+y+z=0 ?$ or am I asking a stupid question. I mean by letting $\displaystyle x+y+z=0$, $\displaystyle x^3+y^3+z^3$ has to be $\displaystyle 3xyz$ but does that mean that $\displaystyle x^3+y^3+z^3$ is generally equal to $\displaystyle 3xyz$.
    Last edited by DIOGYK; Aug 27th 2012 at 10:20 PM.
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    Re: Factoring

    Quote Originally Posted by DIOGYK View Post
    So I guess the answer is $\displaystyle 3(a-b)(b-c)(c-a)$?

    But isn't $\displaystyle x^3+y^3+z^3$ equal to $\displaystyle 3xyz$ only if $\displaystyle x+y+z=0 ?$ or am I asking a stupid question. I mean by letting $\displaystyle x+y+z=0$, $\displaystyle x^3+y^3+z^3$ has to be $\displaystyle 3xyz$ but does that mean that $\displaystyle x^3+y^3+z^3$ is generally equal to $\displaystyle 3xyz$.
    Simplify (a-b) + (b-c) + (c-a). What do you get?
    Thanks from DIOGYK
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  5. #5
    Junior Member DIOGYK's Avatar
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    Re: Factoring

    Quote Originally Posted by awkward View Post
    Simplify (a-b) + (b-c) + (c-a). What do you get?
    0 of course, now I get it. I attacked this problem by fists, filled 2-3 papers haha, and you attacked it with your brain, you're awesome. Thanks for help awkward.
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