Mathematically derive the roots of #color(white)("d")y=x^3-3x-1=0# ?

Given:

#x^3-3x-1 = 0#

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=0#, #c=-3# and #d=-1#, so we find:

#Delta = 0+108+0-27+0 = 81#

Since #Delta > 0# this cubic has #3# Real zeros.

Trigonometric substitution

Since this cubic has #3# real zeros, Cardano's method will result in expressions involving irreducible cube roots of complex numbers. Cardano's method is not wrong, but it is not very friendly, unless the cube roots have a simple form.

As an alternative in such cases, I would choose to use a trigonometric substitution.

Let:

#x = k cos theta#

The trick is to choose #k# such that the resulting expression contains #4 cos^3 theta - 3 cos theta = cos 3 theta#.

We have:

#0 = x^3-3x-1#

#color(white)(0) = k^3 cos^3 theta - 3k cos theta - 1#

#color(white)(0) = k(k^2 cos^3 theta - 3 cos theta) - 1#

#color(white)(0) = 2(4 cos^3 theta - 3 cos theta) - 1" "# with #k=2#

#color(white)(0) = 2cos 3theta - 1#

So:

#cos 3 theta = 1/2#

So:

#3 theta = +-pi/3+2npi" "# for any integer #n#

So:

#theta = +-pi/9+(2npi)/3" "# for any integer #n#

This will give #3# distinct possible values of #x = k cos theta#...

#x = 2 cos theta = 2 cos(pi/9+(2npi)/3)" "# for #n = 0, 1, 2#.