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Math Help - Help! How to foil this cube expression??

  1. #1
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    Help! How to foil this cube expression??

    Consider:

    (x+y)^2

    = (x+y)(x+y)

    = x^2 + xy + xy + y^2

    = x^2 +2xy + y^2

    Question:

    (-2-5)^-3

    = 1/(-2-5)^3

    = 1/(-2-5)(-2-5)(-2-5)

    = 1/?

    How to foil such an expression?

    (x+y)^3

    = (x+y)(x+y)(x+y)

    = ?

    As again, thanks for the help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shenton View Post
    Consider:

    (x+y)^2

    = (x+y)(x+y)

    = x^2 + xy + xy + y^2

    = x^2 +2xy + y^2

    Question:

    (-2-5)^-3

    = 1/(-2-5)^3

    = 1/(-2-5)(-2-5)(-2-5)

    = 1/?

    How to foil such an expression?

    (x+y)^3

    = (x+y)(x+y)(x+y)

    = ?

    As again, thanks for the help.
    You can either look it up as a formula or do it in steps. As you said:

    (x + y)^2 = x^2 + 2xy + y^2. Thus:

    (x + y)^3 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)

    = x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3

    So
    (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

    Similarly:
    (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

    -Dan
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  3. #3
    MHF Contributor Quick's Avatar
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    think of the distributive property:

    x(a+b)=ax+bx

    now try: (a+b)(x+y)

    use distributive property: x(a+b)+y(a+b)

    use distributive property again: ax+bx+ay+by


    now try: (x+y)^3

    rewrite: (x+y)(x+y)(x+y)

    use foil: (x^2+2xy+y^2)(x+y)

    use distributive property: x^2(x+y)+2xy(x+y)+y^2(x+y)

    can you solve from there?
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  4. #4
    Forum Admin topsquark's Avatar
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    With a little work you can find the following:
    (x+y)^2 = x^2 + 2xy + y^2

    (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

    (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

    (x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

    etc.

    Note that in the expansion for (x+y)^n each term takes the form cx^a*y^b with a+b = n, and all possible terms are listed.

    As far as the constants are concerned there are various ways to produce them, the simplest being the combinitorial function, usually written as

    nCr = (n!)/[(r!)(n - r)!]

    You can also use Pascal's Triangle. (Which, if I may be allowed to brag, I discovered on my own. )

    So, to write out, say, (x+y)^4 we do the following:

    (x+y)^4 = (4C0)*x^4*y^0 + (4C1)x^(4-1)*y^1 + (4C2)x^(4-2)*y^2 + (4C3)x^(4-3)*y^3 + (4C4)x^(4-4)*y^4

    (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post

    (x + y)^2 = x^2 + 2xy + y^2. Thus:

    (x + y)^3 = (x + y)(x^2 + 2xy + y^2) = x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)

    = x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3

    So
    (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

    Similarly:
    (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
    Thanks for the step by step proof. That is very helpful in understanding.
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  6. #6
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    Quote Originally Posted by Quick View Post

    now try: (x+y)^3

    rewrite: (x+y)(x+y)(x+y)

    use foil: (x^2+2xy+y^2)(x+y)

    use distributive property: x^2(x+y)+2xy(x+y)+y^2(x+y)
    Thanks again. I think I can solve from there.
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