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Thread: Constructible Solutions: how to prove x^3−6x+2√(pi) has no constructible solutions?

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  1. #1
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    Constructible Solutions: how to prove x^3−6x+2√(pi) has no constructible solutions?

    We know that for cubic polynomial, if there is a constructible root, there is a rational root. But how to prove [tex]x^3-6x+2\sqrt{\pi}[/tex] has no constructible solutions? Any help is really appreciated.
    Last edited by Subhotosh Khan; 12-06-2018 at 09:59 AM.

  2. #2
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    Quote Originally Posted by JojoK View Post
    We know that for cubic polynomial, if there is a constructible root, there is a rational root. But how to prove [tex]x^3-6x+2\sqrt{\pi}[/tex] has no constructible solutions? Any help is really appreciated.
    What are the properties of a constructible root?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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