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Verifying Binomials

  1. Mar 30, 2014 #1

    I was verifying that [tex]\\x^2-y^2=(x-y)(x+y) \\x^3-y^3=(x-y)(x^2+xy+y^2)[/tex] and I realized that can there is a formulation more general like the theorem binomial... my question is: exist a general theorem for sum or difference of terms^n ?
    Last edited: Mar 30, 2014
  2. jcsd
  3. Mar 30, 2014 #2


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    Gold Member

    You can always proceed with the division of x^n-y^n by x-y.
    Then you can guess and then demonstrate what the general solution is.

    For example:

    (x^11 - y^11)/ (x - y) =
    x^10 + x^9 y + x^8 y^2 + x^7 y^3 + x^6 y^4 + x^5 y^5 + x^4 y^6 + x^3 y^7 + x^2 y^8 + x y^9 + y^10

    Try to be specific and by considering multiple examples, you can often find the path to a generalization.
    Never try to be general too early.
  4. Mar 30, 2014 #3


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    Homework Helper
    Gold Member

    (xn - yn) = (x - y)(xn-1 + xn-2y + xn-3y2 + ... + yn-1)

    is considered fairly elementary, but often useful. It is fairly easy to see it is true if you just multiply the x of the first bracket by the second bracket on one line and -y from the first bracket by the second bracket on the second line you will see.

    A connection you should not fail to observe is that this gives you the answer to getting the sum of a geometric series which is

    1 + x + x2 + xn-1

    (I have made the final term xn for easy comparison, but you you'll be able to see what the sum is if the final term is xn).

    The most useful of all applications of this is when x < 1 and n is infinite.
  5. Mar 30, 2014 #4
    Very good!

    And which the formula for xn + yn and for xn + yn + zn?
  6. Mar 31, 2014 #5


    Staff: Mentor

    There is a well-known formula for x3 + y3 (= (x + y)(x2 - xy + y2), and formulas for higher odd powers of xn + yn are fairly well known. Hint: one factor is x + y. You can get the other factor by long division.

    If there's a formula for xn + yn + zn I'm not aware of it.
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