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Math Help - Special Factoring

  1. #1
    Mad
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    Special Factoring

    (a - b)^3 - b^3
    [(a - b) - b][(a - b)^2 + (a - b)b + b^2]


    Is that right?
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  2. #2
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    Correct, Mad!

    . . . Nice going!

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  3. #3
    Mad
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    Gah, this one is driving me nuts :

    Factor:

    27a^4 - a
    a(27a^3 - 1)

    Or can I do more?
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  4. #4
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    Quote Originally Posted by Mad View Post
    Gah, this one is driving me nuts :

    Factor:

    27a^4 - a
    a(27a^3 - 1)

    Or can I do more?
    Note that 27a^3 - 1 = (3a)^3 - 1^3, a difference of two perfect cubes ......
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  5. #5
    Mad
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    I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks.



    EDIT: Just in case...

    (a)(27a - 1)(27a^2 + 27a + 1)
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  6. #6
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    Quote Originally Posted by Mad View Post
    I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks.



    EDIT: Just in case...

    (a)(27a - 1)(27a^2 + 27a + 1)
    Should be 3a's, NOT 27a's ......
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  7. #7
    Mad
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    Sorry, I don't understand. I looked back over my notes but I'm kind of confused.
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  8. #8
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    Quote Originally Posted by Mad View Post
    Gah, this one is driving me nuts :

    Factor:

    27a^4 - a
    a(27a^3 - 1)

    Or can I do more?
    27a^3 - 1 = (3a)^3 - (1)^3

    and
    x^3 - y^3 = (x - y)(x^2 + xy + y^2)

    So let x = 3a and y = 1

    Thus
    27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)

    = (3a - 1)(9a^2 + 3a + 1)

    -Dan
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  9. #9
    Mad
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    Oh, now I understand, thank you.
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