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Math Help - A factorisation question

  1. #1
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    Question A factorisation question

    Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
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  2. #2
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    Hello, anshulbshah!

    I expanded it, then repeatedly factored "by grouping".
    If there is a more direct way, I hope someone finds it.


    Factor: . (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
    b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4

    . . = \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3

    . . = \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)

    . . =\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c)

    . . =\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg]

    . . =\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]

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

    . . =\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg]

    . . = \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg]

    . . =\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]

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

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

    . . =\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]

    . . =\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]


    Answer: . {\color{blue}(a-b)(b-c)(c-a)(a+b+c)}

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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Soroban View Post
    Hello, anshulbshah!

    I expanded it, then repeatedly factored "by grouping".
    If there is a more direct way, I hope someone finds it.

    b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4

    . . = \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3

    . . = \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)

    . . =\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c)

    . . =\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg]

    . . =\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]

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

    . . =\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg]

    . . = \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg]

    . . =\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]

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

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

    . . =\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]

    . . =\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]


    Answer: . {\color{blue}(a-b)(b-c)(c-a)(a+b+c)}
    Wow!
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  4. #4
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    Hello anshulbshah
    Quote Originally Posted by anshulbshah View Post
    Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
    Having seen Soroban's answer, I suppose this is being wise after the event, but here's a quicker method, if you spot it!

    If we put a = b in the expression, we get:

    (b-c)(b^3+c^3) + (c-b)(c^3+b^3) + 0 = 0

    so (a-b) is a factor.

    Similarly (b-c) and (c-a) are also factors.

    Since the expression is of degree 4 in a,b,c, the remaining factor is linear, and by symmetry is therefore k(a+b+c) for some constant k.

    So (b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = k(a-b)(b-c)(c-a)(a+b+c)

    Compare coefficients of, say, bc^3: 1 = k (taking the -b from the first factor, the -c from the second, the c from the third and the c from the fourth)

    \Rightarrow (b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = (a-b)(b-c)(c-a)(a+b+c)

    Or did I cheat a bit?

    Grandad

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  5. #5
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    great work soroban and grandad, whew!

    = -(a - b)(a - c)(b - c)(a + b + c).

    as was observed by grandad, a = b = c. But if you let a = -(b + c); the expression vanishes too.

    Then from a = -(b + c), we have (a + b + c) as a factor also.
    Last edited by pacman; October 1st 2009 at 09:04 AM.
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