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Math Help - Factoring cubes

  1. #1
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    Factoring cubes

    Hi guys. I've been wondering if you could possibly help me. I don't know what to do with that 27 on the second term.
    Attached Thumbnails Attached Thumbnails Factoring cubes-math-problem-3.7000-73.jpg  
    Last edited by Alienis Back; September 29th 2008 at 06:48 AM. Reason: Forgot to attach the problem
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  2. #2
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    Think of it as (4)^3 + (3a)^3
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  3. #3
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    That's really useful...
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  4. #4
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    Quote Originally Posted by Alienis Back View Post
    That's really useful...
    Yes it is.

    Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a...

    4^3 + x^3 = (4+x)(16-4x-x^2)
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  5. #5
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    Quote Originally Posted by Chop Suey View Post
    Yes it is.

    Maybe I should spell it out for you: you should treat 3a as one. Substitute x = 3a...[/tex]
    That is a good idea.

    Quote Originally Posted by Chop Suey View Post
    4^3 + x^3 = (4+x)(16-4x-x^2)
    But I think here there is a mistake. Look at the formula for the sum of cubes above.
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  6. #6
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    Quote Originally Posted by Alienis Back View Post
    That is a good idea.



    But I think here there is a mistake. Look at the formula for the sum of cubes above.
    Yes, you are right. I made a typo there. it should be +x^2 rather than -x^2.
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