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Math Help - binomial expansion help.

  1. #1
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    binomial expansion help.

    Expand (2+y) 3. Hence or otherwise, write down the expansion of (2+xx2) 3 in ascending powers of x.

     (2+y)^3 = 8 + 12y + 6y^2 + y^3

     y = x-x^2

     (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3

     (2+x-x^2)^3 = 8 + 12x(1-x) + 6x(1-x)^2 + x(1-x)^3

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

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

    (2+x-x^2)^3 = 8+ 12x(1-x) + 6x(1-2x+x^2) + x( 1-3x+3x^2- x^3)

     (2+x-x^2)^3 = 8+ 12x-12x^2 + 6x-12x^2+ 6x^3 + x-3x^2 -x^4

    My final answer after simplifying is not correct, can anybody show me where I went wrong, or if this is even a correct method?

    Thanks!

    the correct answer is  (2+x-x^2)^3 = 8+ 12x -6x^2-11x^3 +3x^4 +3x^5-x^6
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  2. #2
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    Quote Originally Posted by Tweety View Post
     (2+y)^3 = 8 + 12y + 6y^2 + y^3

     y = x-x^2

     (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3

     (2+x-x^2)^3 = 8 + 12x(1-x) + 6x(1-x)^2 + x(1-x)^3

    ...
    You've skipped one step:


     (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3


     (2+x-x^2)^3 = 8 + 12(x(1-x)) + 6(x(1-x))^2 + 1(x(1-x))^3

     (2+x-x^2)^3 = 8 + 12x(1-x) + 6x^2(1-x)^2 + 1x^3(1-x)^3
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