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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. –  lhf Sep 17 '12 at 20:56
    
Yes, sorry- will adjust. –  Alyosha Sep 19 '12 at 18:59

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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? –  Alyosha Sep 20 '12 at 18:51
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@Alyosha, not really, sorry. –  lhf Sep 20 '12 at 20:44
    
Oh well. I'll return to this problem in a year or so. –  Alyosha Sep 21 '12 at 17:07
    
@Alyosha, see math.stackexchange.com/questions/152818/… –  lhf Sep 21 '12 at 18:07
    
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? –  Alyosha Sep 27 '12 at 19:47

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