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Thread: a simply math problem-a,b,c

  1. #1
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    Question a simply math problem-a,b,c

    I am writting here because I have small math problem.
    I have to find all configurations of a,b,c (actual numbers) for
    ab(a+b)=bc(b+c)=ca(c+a)

    Have a nice day!
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  2. #2
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    Re: a simply math problem-a,b,c

    $a b(a+b)-c a(c+a)=0$ can be factored

    $a (b-c) (a+b+c)=0$

    similarly for the other pairs of equations

    $b (c-a) (a+b+c)=0$

    $c(a-b)(a+b+c)=0$

    so $a+b+c=0$ gives one set of solutions.
    There are others.
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  3. #3
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    Re: a simply math problem-a,b,c

    Thank You
    Could You explain what happened here...?
    ab(a+b)−ca(c+a)=0 can be factored
    ...?
    a(b−c)(a+b+c)=0
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  4. #4
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    Re: a simply math problem-a,b,c

    $$a b(a+b)-c a(c+a)=a(b(a+b)-c(c+a))=a\left(b a +b^2-c^2-c a\right)=a(a(b-c)+(b-c)(b+c))=a(b-c)(a+b+c)$$
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