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The discriminant for the cubic equation $ax^3+bx^2+cx+d=0$ is

$Δ​\:=b^2c^2−4ac^3−4b^3d−27a^2d^2+18abcd$

And I am aware that you can determine the number of roots a cubic has using method shown below -

$Δ​\:>0$: the equation has three distinct real roots

$Δ​\:=0$: the equation has a repeated root and all its roots are real

$Δ​\:<0$: the equation has one real root and two non-real complex conjugate roots

But I was wondering if one could determine whether a cubic has rational or integer roots, as you can do with the discriminant for quadratics, and if so what the method would be.

I have noticed that with the cubics I have checked: if the discriminant is a perfect square there are 3 integer solutions, although I have not checked many and I am not sure of the reasoning behind it.

Any help would be greatly appreciated.

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  • $\begingroup$ This is a good question, but I think you meant something else when you wrote “cubic for quadratics” $\endgroup$ Commented May 19, 2020 at 17:49
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    $\begingroup$ Even if you use the discriminant for quadratics, you cannot say wether the roots are integer or rational.. You also need to know the coefficients $\endgroup$
    – Exodd
    Commented May 19, 2020 at 18:01
  • $\begingroup$ Yes sorry you are right I meant to say discriminant for quadratics $\endgroup$
    – user578923
    Commented May 19, 2020 at 20:12
  • $\begingroup$ And I meant just rational not integer or rational. $\endgroup$
    – user578923
    Commented May 19, 2020 at 20:13

1 Answer 1

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When a monic cubic has square discriminant but no rational roots, what we expect is real roots that can be written as (doubled) cosines, or sums of them.

$$ x^3 + x^2 - 2x - 1 $$ has $$ 2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \; 2 \cos \frac{8 \pi}{7} \; , \; \; $$

more in a minute $$ x^3 - 3x + 1 $$ has $$ 2 \cos \frac{2 \pi}{9} \; , \; \; 2 \cos \frac{4 \pi}{9} \; , \; \; 2 \cos \frac{8 \pi}{9} \; , \; \; $$ $$ $$ $$ x^3 + x^2 - 4x + 1 $$ has $$ 2 \cos \frac{2 \pi}{13} + 2 \cos \frac{10 \pi}{13}\; , \; \; 2 \cos \frac{4 \pi}{13} + 2 \cos \frac{6 \pi}{13} \; , \; \; 2 \cos \frac{8 \pi}{13} +2 \cos \frac{12 \pi}{13} \; , \; \; $$

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  • $\begingroup$ Thank you for the response, is there any way you can tell if the equation has rational roots if you are just given the discriminant value? As I understand your point but you need the original equation as well. $\endgroup$
    – user578923
    Commented May 19, 2020 at 20:17
  • $\begingroup$ The representation of roots as trig. values is brilliant. $\endgroup$
    – NoChance
    Commented Feb 6, 2023 at 3:20

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