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Math Help - Sum or difference of cubes.

  1. #1
    Member integral's Avatar
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    Sum or difference of cubes.

    a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})
    And just switch the sign after a^3-1 for  a^3-b^3

    My question:
    I have read that the last two signs are constant (never changing) Unless the initial sign is different. I.e.
    a-b =  +,+
    a+b =  -,+

    now take this example:

    8x^3-1
    when worked out it equals:
    (\sqrt [3] {8x^3} + \sqrt [3] {-1})(8x^{3-1}+(-1)2x+(-1^{3-1})
    (2x-1)(4x^2-2x-1)

    The end signs are clearly not  +,+
    Am I missing something?
    Last edited by integral; January 14th 2010 at 05:13 PM. Reason: equation typo xD
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  2. #2
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    Quote Originally Posted by integral View Post
    a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})
    And just switch the sign after a^3-1 for  a^3-b^3

    My question:
    I have read that the last two signs are constant (never changing) Unless the initial sign is different. I.e.
    a-b =  +,+
    a+b =  -,+

    now take this example:

    8x^3-1
    when worked out it equals:
    (\sqrt [3] {8x^3} + \sqrt [3] {-1})(8x^{3-1}+(-1)2x+(-1^{3-1})
    (2x-1)(4x^2-2x-1)

    The end signs are clearly not  +,+
    Am I missing something?
    a^3 \pm b^3 = (a - b)(a^2 \mp ab + b^2).


    8x^3 - 1 = (2x)^3 - (1)^3

     = (2x - 1)(4x^2 + 2x + 1).
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  3. #3
    Member integral's Avatar
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    a^{3}+b^{3}=( \sqrt[3]{a^{3}}+ \sqrt[3]{b^{3}}) (a^{3-1} -\sqrt[3]{a^3b^3} +b^{3-1})

    8x^3-1=( \sqrt[3]{8x^{3}}+ \sqrt[3]{-1^{3}}) (8x^{3-1} -\sqrt[3]{8x^3b^3} +(-1^{3-1}))
    8x^3-1=(2x-1)(4x^2+(-1)2x+(-1)<br />
    8x^3-1=(2x-1)(4x^2-2x-1)<br />
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  4. #4
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    What you have at the moment is a DIFFERENCE of cubes.

    Use the Difference of Two cubes rule.

    a^3 - b^3 = (a - b)(a^2 + ab + b^2).

    See my post above.


    Edit: Also, -\sqrt[3]{-1(8x^3)} = -\sqrt[3]{-8x^3} = -(-2x) = 2x


    and + (-1)^2 = +(1) = 1.


    So you do get +, + as your final two signs.
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