Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum/Diff. formulas

  1. Mar 1, 2005 #1
    For the difference of two squares you can factor using: [tex]a^2 - b^2 = (a + b)(a - b)[/tex].

    And for the sum and difference of two cubes:
    [tex]a^3 + b^3 = (a+b)(a^2 - ab + b^2)[/tex]
    [tex]a^3 - b^3 = (a-b)(a^2 + ab + b^2)[/tex]

    How are these formulas found...?
    Can it be done with basic algebra (am I missing something really simple)?
    For example, it is possible to take [tex]ax^2 + bx + c = 0[/tex] and solve for x to get the quadratic formula. But how do you rework [tex]a^2 - a^2[/tex], or [tex]a^3 \pm b^3[/tex] into their respective formulas? What about with higher exponents (i.e. [tex]a^4 \pm b^4[/tex] or or [tex]a^5 \pm b^5[/tex])?

    I realize that the resulting factors must, when multiplied, have middle terms that will completely cancel each other out leaving only [tex]a^n \pm b^n[/tex], but how do you work backwards to find the middle terms of the factors?

    Last edited: Mar 1, 2005
  2. jcsd
  3. Mar 1, 2005 #2
    You go the other way, i.e.
    [tex] a^{2}-b^{2}=a^{2}-ab+ab-b^{2}=a(a-b)+b(a-b)=(a-b)(a+b) [/tex]
    [tex] a^{3}-b^{3}=a^{3}+a^{2}b-a^{2}b+ab^{2}-ab^{2}-b^{3}=a^{2}(a-b)+ab(a-b)+b^{2}(a-b)=(a-b)(a^{2}+ab+b^{2}) [/tex]...
    Last edited: Mar 1, 2005
  4. Mar 1, 2005 #3
    I understand that you can take the formula and work it backwards to find the middle terms. But what if you don't have the formula? Someone had to find the formula, so how did they find it?

    Forgive me if I'm being obtuse, but what mathematical process is used to get from [tex]a^3 - b^3[/tex] to [tex]a^3-a^2b+ab^2+a^2b-ab^2+b^3[/tex]

    I'm working under the assumption that the end formula is not necessary in mathematically determining the middle terms... am I incorrect?
    I ask because without the formula as a reference, it doesn't seem like it is possible to find the middle terms without a bit of trial and error. Is this the case? And what about cases where the binomial can't be factored? (i.e. [tex]a^2 + b^2[/tex]).

    Last edited: Mar 1, 2005
  5. Mar 1, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Did u ever hear of polynomials' division...?The polynomial [tex] P(a)=a^{n}-b^{n} [/tex] has the root a=b...Agree.Then please divide the 2 polynomials...

    As for his brother,[tex] P(a)=a^{n}+b^{n} [/tex],well,it can't be factored into the reals...Not really for doing integrals in the reals.

  6. Mar 1, 2005 #5
    Last edited: Mar 1, 2005
  7. Mar 1, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    Hold it,what have Binomial numbers (i hope they make up Pascal's triangle) with [itex] a^{n}-b^{n} [/tex]...?Binomial numbers come from [itex] (a+b)^{n} [/itex],which totally another animal...:wink:

  8. Mar 1, 2005 #7
    Creativity aided by experience.
  9. Mar 1, 2005 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    There is NO process that will get you from [tex]a^3 - b^3[/tex] to [tex]a^3-a^2b+ab^2+a^2b-ab^2+b^3[/tex]

    Perhaps you are thinking of
    [tex]a^3- b^3= (a- b)(a^2+ ab+ b^2)[/tex]

    The way that was found was exactly what you were told before: people noticed that multiplying (a-b)(a2+ ab+ b2 gave you
    a3- b3

    It is true that an- bn= (a-b)(an-1+ an-2b+ ... + abn-1+ bn.

    IF n IS ODD then an+ bn= (a+b)(an-1- an-2b+ ... - abn-1+ bn. But an+ bn for n even, such as a2+ b2 or a4+ b4 cannot be factored (in terms of real numbers).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Sum/Diff. formulas
  1. Trig sum formulas (Replies: 4)