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Factoring the Sum of Two Fifth Powers

  1. Jun 18, 2008 #1
    My problem is to factor [tex]x^{5}[/tex] + [tex]y^{5}[/tex].

    I get [tex](x+y)(x^{8}-x^{4}y^{4}+y^{8})[/tex].

    However, the answer is [tex](x+y)(x^{4}-x^{2}y^{2}+y^{4})[/tex].
  2. jcsd
  3. Jun 18, 2008 #2


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    Neither is right! Multiplying out in either cases does not give you the original back.

    The correct answer is (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
  4. Jun 19, 2008 #3




    What am I doing wrong?


    Your answer:
  5. Jun 19, 2008 #4


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    The first and last steps are the only ones that make any sense; I have no idea what you were trying to do inbetween.
  6. Jun 19, 2008 #5
    How about you try factoring 5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4 instead? This gives 5*xy( (x+y)^3 - x^2*y - x*y^2) = 5*xy( (x+y)^3 - xy(x + y)) = 5*xy*(x+y)*((x+y)^2 - xy). Since x^5 + y^5 = (x+y)^5 - (5*x^4*y + 10*x^3*y^2 + 10* x^2 * y^3 + 5*x*y^4), we obtain x^5 + y^5 = (x+y)^5 - 5*xy*(x+y)*((x+y)^2 - xy) = (x+y)((x+y)^4 - 5*xy*(x+y)*((x+y)^2 - xy)), and take it from there (you can extract (x+y) still).
    Last edited: Jun 19, 2008
  7. Jun 19, 2008 #6

    Gib Z

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    You could think of that expression as [/itex]x^5 - (-y)^5[/itex] and apply that into [tex]a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-1})[/tex]. You can verify that general equation by expanding the RHS.
  8. Jun 19, 2008 #7


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    Your last assertion makes no sense. You have terms x9 and y9 among others. They seem to come out of nowhere!!!!
  9. Jun 19, 2008 #8
    I thought applying the sum of cubes on the cubed power parts would have worked [tex]x^{3}x^{2}+y^{3}y^{2}[/tex].

    Now I realize why that was inappropriate.
    Last edited: Jun 19, 2008
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