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Math Help - Factorisation...

  1. #1
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    Factorisation...

    Hi,

    This expression can factorised completely.

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

    Here is how far I've gotten with this.

    (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)

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

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

    =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b-a).

    Now I need a little push in the right direction. Alternatively, if I've started tackling this wrong a push on where to start it.

    Thanks.
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  2. #2
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    Quote Originally Posted by bluntpencil View Post
    Hi,

    This expression can factorised completely.

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

    Here is how far I've gotten with this.

    (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)

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

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

    =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b-a).

    Now I need a little push in the right direction. Alternatively, if I've started tackling this wrong a push on where to start it.

    Thanks.
    The final step should have been

    =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b\color{red}+\color{black}a)
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  3. #3
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    Argh, damn and blast it. I wrote it out correctly but type it in wrong.

    The solution to this is (a-b)(b-c)(c-a)(a+b+c). As you can see I'm alittle bit off of that at this time.

    Thanks.
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