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Math Help - Factors Question

  1. #1
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    Factors Question

    I have a really hard question from IMO (1984) it goes:
    Find the factors of a^3(b-c) + b^3(c-a) + c^3(a-b). Does anyone know how to solve this? If possible, show working/reasoning as well? Thanks
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  2. #2
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    Hello, cedricc!

    We will factor "by grouping" ... repeatedly.


    Factor: . a^3(b-c) + b^3(c-a) + c^3(a-b)

    We have: . a^3b - a^3c + b^3c - ab^3 + ac^3 - bc^3

    Rearrange: . a^3b - a^3c - ab^3+ac^3 + b^3c-bc^3

    Factor: . a^3(b-c) - a(b^3-c^3) + bc(b^2-c^2)

    Factor: . a^3(b-c) - a(b-c)(b^2 + bc + c^2) + bc(b-c)(b+c)

    Factor: . (b-c)\bigg[a^3 - a(b^2+bc+c^2) + bc(b+c)\bigg]

    Simplify: . (b-c)\bigg[a^3 - ab^2 - abc - ac^2 + b^2c + bc^2\bigg]

    Rearrange: . (b-c)\bigg[b^2c - ab^2 + bc^2 - abc - ac^2 + a^3\bigg]

    Factor: . (b-c)\bigg[b^2(c-a) + bc(c-a) - a(c^2-a^2)\bigg]

    Factor: . (b-c)\bigg[b^2(c-a) + bc(c-a) - a(c-a)(c+a)\bigg]

    Factor: . (b-c)(c-a)\bigg[b^2 + bc - a(c+a)\bigg]

    Simplify: . (b-c)(c-a)\bigg[b^2 + bc - ac - a^2\bigg]

    Rearrange: . (b-c)(c-a)\bigg[b^2-a^2 + bc - ac\bigg]

    Factor: . (b-c)(c-a)\bigg[(b-a)(b+a) + c(b-a)\bigg]

    Factor: . (b-c)(c-a)(b-a)\bigg[b + a + c\bigg]


    Therefore: . (a-b)(b-c)(a-c)(a+b+c)



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  3. #3
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    Thank you so much, Soroban! =]
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  4. #4
    Super Member Bacterius's Avatar
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    Woah Soroban, thumbs up for having kept the equation valid to the end ... I'd probably have made a mistake at the seventh line
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