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Factoring problem

  1. Nov 20, 2006 #1
    How do I factor sin(^6)x + cos(^6)x?
     
  2. jcsd
  3. Nov 20, 2006 #2
    Use x^3+y^3=(x+y)(x^2-xy-y^2).
     
  4. Nov 21, 2006 #3

    dextercioby

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    [tex] \sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.[/tex]

    Daniel.
     
  5. Nov 21, 2006 #4
    Doesn't [tex] ( \sin^{2} x +\cos^{2} x ) ( \sin^{4}x +\cos^{4} x ) - 2 \sin^{2}x \cos^{2} x ( \sin^{2} x +\cos^{2} x) = \sin^6 x+ \cos^6 x - \sin^2 x \cos^4 x- \sin^4 x \cos^2 x [/tex]?:confused:
     
    Last edited by a moderator: Nov 21, 2006
  6. Nov 21, 2006 #5

    arildno

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    Dearly Missed

    Whereby the hitherto unknown equality:
    [tex]\frac{1}{8}+\frac{1}{8}=0[/tex]
    is proven. :smile:
     
  7. Nov 22, 2006 #6

    dextercioby

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    Yes, thought it was 2 simple 2 be true. You can drop that 2, then. :d

    Daniel.
     
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