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Math Help - Three-equations twelfth-grade system

  1. #1
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    Three-equations twelfth-grade system

    * x^4 = y + z
    * y^4 = x + z
    * z^4 = x + y

    I'd truly appreciate a pointer or a suggestion, even a broad one.

    Thanks in advance!
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  2. #2
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    Re: Three-equations twelfth-grade system

    Quote Originally Posted by AndrewTahoe View Post
    * x^4 = y + z
    * y^4 = x + z
    * z^4 = x + y

    I'd truly appreciate a pointer or a suggestion, even a broad one.

    Thanks in advance!
    Subtract the second equation from the first to get x^4- y^4= y- x. Factor the left side.
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    Re: Three-equations twelfth-grade system

    Thank you very much!

    Yes, I had tried it before. You get:

    x^3 + xy^2 + x^2*y + y^3 + 1 = 0

    What about now? I tried to obtain two lower-grade factors with Ruffini's rule, but I can't find a root for the expression.
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    Re: Three-equations twelfth-grade system

    Don't forget that you divided by x- y to get that. So x= y is a separate solution. If x= y, what must z be?

    Also, if you subtract the third equation from the first, x- z= 0 or [itex]x^3+ zx^2+ xz^2+ z^4+ 1= 0[/itex]. Use that with your equation in x and y.
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    Re: Three-equations twelfth-grade system

    Sorry, I don't get it. When you divide by x-y, you state that x-y != 0 -> x != y. How can x=y be a solution?

    Also, can you be a bit more specific in the other approach?

    Thanks a lot for your attention!
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  6. #6
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    Re: Three-equations twelfth-grade system

    Quote Originally Posted by HallsofIvy View Post
    Subtract the second equation from the first to get x^4- y^4= y- x. Factor the left side.
    When you divide by y - x you do not get 0 on the RHS, you get 1. I think this is causing some of your confusion.

    Quote Originally Posted by AndrewTahoe View Post
    Sorry, I don't get it. When you divide by x-y, you state that x-y != 0 -> x != y. How can x=y be a solution?

    Also, can you be a bit more specific in the other approach?

    Thanks a lot for your attention!
    x -y is a solution, but not the only solution.

    Let's try factoring instead of division:

    First let's put things in an easier format:
    x^4 - y^4 = y - x

    (x^2 - y^2)(x^2 + y^2) = y - x

    (x - y)(x + y)(x^2 + y^2) = y - x

    Note the change I'm making on the RHS:
    (x - y)(x + y)(x^2 + y^2) = -(x - y)

    (x - y)(x + y)(x^2 + y^2) + (x - y) = 0

    Now factor:
    (x - y) \left [ (x + y)(x^2 + y^2) + 1 \right ] = 0

    So we know that either
    x - y = 0

    or
    (x + y)(x^2 + y^2) + 1 = 0

    The first equation gives y = x. The other equation? I'm not telling. Give it a try.

    -Dan
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    Re: Three-equations twelfth-grade system

    Thanks a lot! I'll give it a try.
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