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Math Help - quick question about cubing polynomials.

  1. #1
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    quick question about cubing polynomials.

    I just want to understand how you get the three when you have to cube three binomials together.

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

    my tutor gave me this formula to follow when i come across anything I have to cube. However I'm not sure how she got the 3 there.
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    Quote Originally Posted by sara213 View Post
    I just want to understand how you get the three when you have to cube three binomials together.
    (a+b)^3=a^3+b^3+3a^2b+3ab^2
    have to cube. However I'm not sure how she got the 3 there.
    (a+b)(a+b)(a+b)=(a)(a)(a)+(a)(a)(b)+(a)(b)(a)+(a)(  b)(b)+\cdots
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    Quote Originally Posted by Plato View Post
    (a+b)(a+b)(a+b)=(a)(a)(a)+(a)(a)(b)+(a)(b)(a)+(a)(  b)(b)+\cdots
    Thanks but that doesn't answer my question, i know that part already.

    I'm talking about was the actually about the 3 in front of a^2b
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    Quote Originally Posted by sara213 View Post
    Thanks but that doesn't answer my question, i know that part already. I'm talking about was the actually about the 3 in front of a^2b
    I beg your pardon. It actually does answer your question.
    If you follow how to multiply then you know that a^2b comes from (a)(a)(b)+(a)(b)(a)+(b)(a)(a)=3a^2b.
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    Quote Originally Posted by sara213 View Post
    Thanks but that doesn't answer my question, i know that part already.

    I'm talking about was the actually about the 3 in front of a^2b
    His point is that there are terms (a)(a)(b), (a)(b)(a), and (b)(a)(a) that come up. Since there are three of them, you get a 3 in front of the a^2b.

    However let's look at this differently. (a + b)^2 = a(a + b) + b(a + b), right? Then you expand and simplify to get (a + b)^2 = a^2 + 2ab + b^2. The 2 in front of the ab appears because we have a term ab and ba.

    Similarly:
    (a + b)^3 = (a + b)(a + b)^2 = (a + b)(a^2 + 2ab + b^2)

    = a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)

    Expand this and simplify. Here you will find, for instance, that you have a 2a^2b and a ba^2 which gives you 3a^2b.

    -Dan
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    The order doesn't matter in multiplication - 5*3 is the same as 3*5. This extends to letters too so (a)(a)(b) = (a)(b)(a) = (b)(a)(a) and since you have three lots in your expansion it gives 3a^2b

    If "proving" it isn't all that important look up Pascal's triangle - that gives you the coefficients necessary. Or, better yet, the binomial theorem
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  7. #7
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    Okay got it...sorry plato, I did not understand what you are trying to say in the first post.
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