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Help with Binomial Expansion question.

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    Hi, I did this a few weeks ago, but now I have come back to it I feel I am doing the right thing but I am getting the wrong answer everytime I try it.

    Expand (1 + x + x²)^3 as far as the x^3 term.

    I did this: [ 1 + (x + x²)]^3

    = 1 + [x^3] + [3x²(x²)] + [3x(x²)²] + (x²)^3

    And when I simplify this (excluding the terms with powers greater than 3), I get the wrong answer. Have I done something wrong or is the answer wrong? I would give you the answer so you can tell me if its right, but my book isn't currently with me, sorry.
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    (Original post by ERdoctor)
    Hi, I did this a few weeks ago, but now I have come back to it I feel I am doing the right thing but I am getting the wrong answer everytime I try it.

    Expand (1 + x + x²)^3 as far as the x^3 term.

    I did this: [ 1 + (x + x²)]^3

    = 1 + [x^3] + [3x²(x²)] + [3x(x²)²] + (x²)^3

    And when I simplify this (excluding the terms with powers greater than 3), I get the wrong answer. Have I done something wrong or is the answer wrong? I would give you the answer so you can tell me if its right, but my book isn't currently with me, sorry.
    Where did you use binomial expansion here?
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    SORRY!

    I think I know what i did wrong. I should have factorised the (x + x²)^3 bracket to get a 1 for the first term, other wise the expansion won't work. My mistake!

    Am I right?
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    Have you multiplied the terms by the binomial coefficients as well? You know, the ones you get off Pascal's triangle?
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    (Original post by raheem94)
    Where did you use binomial expansion here?
    I didn't show my full working out, but I did use binomial expansion. I think I know where I got it wrong though, I should have factorised more and got a 1 as my first term instead of X.

    So rather than having [1 + (x + x²)]^3

    I would have [1 + x(1 + x)]^3 Before expanding binomially.

    Is that right?
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    (Original post by ERdoctor)
    SORRY!

    I think I know what i did wrong. I should have factorised the (x + x²)^3 bracket to get a 1 for the first term, other wise the expansion won't work. My mistake!

    Am I right?
    Just use the formula,
     1 +nx + \dfrac{n(n-1)}{2!}x^2  + \dfrac{n(n-1)(n-2)}{3!}x^3 +.....

    Where n=3 and x=(x+x^2)
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    Ahhh, so you do two separate binomial expansions, and then add their x^3 terms together... I think that could work.
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    (Original post by ERdoctor)
    I didn't show my full working out, but I did use binomial expansion. I think I know where I got it wrong though, I should have factorised more and got a 1 as my first term instead of X.

    So rather than having [1 + (x + x²)]^3

    I would have [1 + x(1 + x)]^3 Before expanding binomially.

    Is that right?
    You should use, [1 + (x + x²)]^3, i have tried it and i get the right answer.

    Spoiler:
    Show

     [1 + (x + x²)]^3 = 1 + 3(x+x^2) + 3(x+x^2)^2 + (x+x^2)^3

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    (Original post by raheem94)
    Just use the formula,
     1 +nx + \dfrac{n(n-1)}{2!}x^2  + \dfrac{n(n-1)(n-2)}{3!}x^3 +.....

    Where n=3 and x=(x+x^2)
    Yeah, that worked. I knew I was over complicating it again.

    Thanks! My answer was 1 + 3x +6x² + 7x^3
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    (Original post by ERdoctor)
    Yeah, that worked. I knew I was over complicating it again.

    Thanks! My answer was 1 + 3x +6x² + 7x^3
    Your answer is correct .

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