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Math Help - [SOLVED] Simplifying perfect squares

  1. #1
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    [SOLVED] Simplifying perfect squares

    I know for example (x^2-16) = (x+4) (x-4) but what about something like (x^3-64)? I can't find the rule to solve something like this. I think it involves 3 terms but I can't figure it out.
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  2. #2
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    Quote Originally Posted by thekrown View Post
    I know for example (x^2-16) = (x+4) (x-4) but what about something like (x^3-64)? I can't find the rule to solve something like this. I think it involves 3 terms but I can't figure it out.
    Hi,

    For  a^3 \pm b^3

     a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

    This is known as the sum and difference of two cubes.

    In your case, a = x and b = 4
    Last edited by Gusbob; March 8th 2010 at 03:39 PM. Reason: missed a bracket and spelling errors
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  3. #3
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    (x^3-64)

    I keep getting the wrong results.

    (x-4)(x^2 -4x + 16) = x^3 - 8x^2 -64
    (x+4)(x^2 -4x+16) = x^3 + 64)
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  4. #4
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    Be grateful that the problem is as simple as x^3 = 64 (meaning no terms that are of the second or first degree in this equation).

    You can't simplify this type of problem further and you'll have to know from memory or a table that 4^3 = 64.
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