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Math Help - Taking an intiger out of brackets which are cubed

  1. #1
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    Taking an intiger out of brackets which are cubed

    Apparently.... (x+c-1)^3 - (x-1)^3 = 3c(x-1)^2 - 3(x-1)c^2 + c^3


    Someone please explain why or what rule this is, Will be much appreciated

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  2. #2
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    Quote Originally Posted by Snaggle View Post
    Apparently.... (x+c-1)^3 - (x-1)^3 = 3c(x-1)^2 - 3(x-1)c^2 + c^3
    (x+c-1)^3=x^3+3x^2c-3x^2-6xc^2+3x+c^3-3c^2+3c-1

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

    Subtract and factor.
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  3. #3
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    You may also use the difference of two cubes a^3-b^3 = (a-b)(a^2+ab+b^2) but Plato's method is better because of the factors involved when doing the difference of two cubes
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  4. #4
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    A slight variation- if we let a= x- 1 then this is (a+c)^2- a^3= a^3+ 3a^2c+ 3ac^2+ c^3- a^2= 3a^2c+ 3ac^2+ c^3. Now replace a by x- 1 again to get
    3c(a-1)^2+ 3c^2(a- 1)+ c^3
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