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Math Help - Factorisation

  1. #1
    Member classicstrings's Avatar
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    Factorisation

    Hi How would I factorise this:

    a^6 + b^6

    I change it into a cubic so,

    (a^3)^2 + (b^3)^2 then ?

    And how would I fully factorise this: x^3 - 3x^2 - 4

    Thanks!
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  2. #2
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    Quote Originally Posted by classicstrings
    Hi How would I factorise this:
    a^6 + b^6
    ...
    Hello, classicstring,

    I can't offer you a more systematic way to do your problem. I did a little bit trial and error and long division and found this product:

    a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)

    Bye

    EB
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by earboth
    Hello, classicstring,

    I can't offer you a more systematic way to do your problem. I did a little bit trial and error and long division and found this product:

    a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)

    Bye

    EB
    Another way to do it will give you...
    a^6+b^6=-1\left(\left(a^3-b^3\right)\left(b^3-a^3\right)\right)
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  4. #4
    Member classicstrings's Avatar
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    I just realised that I think it is like cubic expansion, just doubling the powers.

    Any ideas on this one?

    x^3 - 3x^2 - 4

    I think there are two nonreal solns and one real one.

    Thanks!
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  5. #5
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    Quote Originally Posted by classicstrings
    I just realised that I think it is like cubic expansion, just doubling the powers.
    Elementry, my Dear classicstrings.
    Use,
    x^3+y^3=(x+y)(x^2-xy+y^2)
    Thus,
    a^6+b^6=(a^2)^3+(b^2)^3
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  6. #6
    Member classicstrings's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Elementry, my Dear classicstrings.
    Use,
    x^3+y^3=(x+y)(x^2-xy+y^2)
    Thus,
    a^6+b^6=(a^2)^3+(b^2)^3
    I like your panache PerfectHacker, the explanation totally clears things up, - EDIT: yeah I was confused a bit as I had wrote

    <br />
a^6+b^6=(a^3)^2+(b^3)^2<br />

    Which is wrong!

    By any chance can you help me with factorising this one?

    x^3 - 3x^2 - 4
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