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Math Help - Factoring the sum of cubes

  1. #1
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    Factoring the sum of cubes

    Hi MHF,

    Usually I can factor, but this particular problem is just killing me today. I have been trying to figure this out for over an hour...

    128r^3+686z^3

    I know two things...

    I know I can immediatly factor out the 2 and take it down to 2(64r^3+343z^3).

    And I know the answer is going to be in the form of 2(binomial)(trinomial).

    But for the life of me I cannot seem to find the proper middle terms to make it work.

    Thanks in advance.
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  2. #2
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    consider 4^3=64 and 7^3 = 343

    now make your problem into the form

    2(a^3+b^3) = 2(a+b)(a^2-ab+b^2)
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  3. #3
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    Quote Originally Posted by pickslides View Post
    consider 4^3=64 and 7^3 = 343

    now make your problem into the form

    2(a^3+b^3) = 2(a+b)(a^2-ab+b^2)
    2(4r+7z)(16r^2-28rz+49z^2)

    Yeah, that should have been obvious. After doing 3 hours of math, even the obvious things become so complicated.

    I think it's time to break and brew some coffee! Thank you Pickslides.
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