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$x=\sqrt[3]{q+\sqrt{q^2-p^3}}-\sqrt[3]{-q+\sqrt{q^2-p^3}}$

Why is it that this must be true $q^2-p^3<0$ for x to have 3 distinct real roots?

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    Do you mean $q^2−p^3<0$? Square roots are always positive.2012-09-17
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    Yes, sorry- will adjust.2012-09-19

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This is the casus irreducibilis of the cubic. You need to use complex numbers even if the roots are all real. The proof needs Galois theory.

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    Thank you. Before I try to delve into it, is learning Galois theory quickly feasible?2012-09-20
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    @Alyosha, not really, sorry.2012-09-20
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    Oh well. I'll return to this problem in a year or so.2012-09-21
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    @Alyosha, see http://math.stackexchange.com/questions/152818/what-is-the-prerequisite-knowledge-for-learning-galois-theory2012-09-21
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    Sorry to whip a dead post, but where I'm reading it says that you can prove this by 'considering the nature of stationary points'. Is this correct?2012-09-27
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    @Alyosha, it may be. My point is that proving that you cannot solve a cubic with three real roots using radicals without using complex numbers needs Galois theory. Not that if $q^2-p^3<0$ then it has three real roots. I seems that I did not really answer your question. I suggest you add an answer yourself.2012-09-27