<s>[INST] Cubic Discriminant Uses 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.
 [/INST] 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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