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Difference of squares

  1. May 17, 2005 #1
    How do I factor this equation
  2. jcsd
  3. May 17, 2005 #2


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    Difference of cubes you mean

    [tex] (x - 1)(x^2 + x + 1) [/tex]
  4. May 17, 2005 #3


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    No cubes there,just squares.

    [tex] \left(x^{\frac{3}{2}}\right)^{2}-1^{2} [/tex]


    P.S.Regarding a proof for the formula Cyclovenom posted,try polynomial division.
  5. May 17, 2005 #4
    Isn't [tex]x \times x^2 = x^3[/tex] a cube? Or is the power of 3 now a square?! :redface:

    Cyclovenom's factorisation [tex] (x - 1)(x^2 + x + 1) [/tex] works fine and the problem is typically the difference of two cubes. Your factorisation works fine too.

    Last edited: May 17, 2005
  6. May 17, 2005 #5
    It does work

    [tex] a^2 - b^2 = (a+b)(a-b) [/tex]

    [tex] a = x^{\frac{3}{2}}, b = 1[/tex]

    [tex] (x^\frac{3}{2} + 1)(x^\frac{3}{2}-1)[/tex]
  7. May 17, 2005 #6
    Hey i was reading something on paper and was thinking in about 5 dimensions and it isn't always easy for me to come back down to one - so sorry.

    I corrected the erratum a few minutes later when I begn thinking about complex numbers and the post now stands happy and correct.

    The mistake i made was not treating the statement as
    [tex]- 1^2[/tex] which equals [tex]-1[/tex]

    but as
    [tex](- 1)^2[/tex] which of course equals [tex] 1[/tex]

    Last edited: May 17, 2005
  8. May 17, 2005 #7
    Well yes [tex]x^3[/tex] is certainly a cube, but Daniel is saying that [tex]\left(x^{\frac{3}{2}}\right)^{2}=x^3[/tex].
  9. May 17, 2005 #8
    z-component I was making reference to

    when he was talking about the difference of two cubes.

    i of course was not trying to deny the fact that [tex]\left(x^{\frac{3}{2}}\right)^{2}=x^3[/tex]

  10. May 17, 2005 #9
    Ah, ok. I understand.
  11. May 17, 2005 #10


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    Read the title of the thread "difference of squares", which would imply it's asking how to factor my difference of squares as apposed to any other method. It's probabily a mistake by the author but still...
  12. May 17, 2005 #11


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    However, to get back to the question posed before dextercioby threw his monkey wrench into the machinery: x3- 1 is a sum of cubes and can be factored as x3- 1= (x-1)(x2+ x+ 1). In general xn- yn= (x-y)(xn-1+ xn-2y+ . . .+ x2yn-3+ xyn-2+ yn-1).
  13. May 18, 2005 #12

    how is that last general form, derived ?
  14. May 18, 2005 #13


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    Polynomial division.You know that the polynomial [itex] P(x)=x^{n}-y^{n} \ ,\mbox{n=odd} [/itex] has at least one real root and it's easy to see that the root is "y".Therefore you know that the remainder of the division of the polynomial [itex] P(x) [/itex] through the monom [itex] x-y [/itex] is zero and all u have to do is compute the ratio.

  15. May 18, 2005 #14
    May I add, lettng [tex]\zeta =\frac{-1+\sqrt-3}{2} [/tex], we can continue the factorization arriving at:

    [tex]X^3-Y^3 = (X-Y)(X-\zeta{Y})(X-\zeta^2Y)[/tex]
    Last edited: May 18, 2005
  16. May 18, 2005 #15


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    Polynomial division is one way. The other is using the formula for sum of a geometric progression.

    Note that the series that HallsofIvy posted is a geometric sum with first term [itex]x^{(n-1)}[/itex] and common ratio [itex]\frac{y}{x}[/itex] giving the sum (to n terms) of :

    [tex]\frac{x^{n-1}({(\frac{y}{x})}^n - 1)}{\frac{y}{x} - 1}[/tex] which reduces to [tex]\frac{y^n - x^n}{y - x}[/tex]
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