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.
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.
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