[I hope I transcribed the formulae correctly.] The usual trig solution for a monic real cubic eqn with 3 real roots has the high school teacher pulling an obscure trig identity out of a hat in deus ex machina fashion, leaving the high school student feeling inadequate, helpless, irritable, and (s)he probably then tunes out of the rest of the presentation & starts texting on his/her cellphone. Here's a version of a cubic solution that a student familiar with complex numbers should feel very good about. We first eliminate the quadratic x^2 term in the usual fashion by shifting the center of mass of the roots to the origin. Finding the center of mass doesn't require solving the equation -- it is merely -1/3 the coefficient of the quadratic term. It is pretty easy to motivate this step -- we almost always translate to the origin whenever easily possible! That having been done, we are left with the so-called "incomplete" cubic: x^3+px+q = 0. Since the 3 roots r1,r2,r3 are real, they can take on any real values, so long as r1+r2+r3=0 (remember, we shifted the center of mass to the origin). All three could be equal (hence zero), or two could be equal (hence equal to minus half the third root). We assume that all three roots are distinct, and will check back later to make sure that everything still works for the other cases. Here is the modest bunny we will pull out of the hat: we conjecture that all three _real_ roots are the _real parts_ of the 3 vertices of an equilateral triangle in the complex plane centered on the origin. We first check the dimensionality: our triangle has only two real parameters: the angle a with the x-axis (real axis) and the radius r of the circumcircle. But our incomplete cubic also has only two real parameters, p,q, so we are still in business. We next realize that the three vertices of the equilateral triangle are the three cube roots of a complex number: z^3 = (r*(cos(a)+i*sin(a)))^3 = r^3*(cos(3a)+i*sin(3a)) [de Moivre] Now let z=x+iy, so z^3 = (x+iy)^3 = (x^3-3xy^2) - i*(y^3-3yx^2) = r^3*(cos(3a)+i*sin(3a)) Since we are projecting onto the x-axis, we focus only on the real part: x^3-3xy^2 = r^3*cos(3a) But how to get rid of y^2 ? We have the equation of the circumcircle: x^2+y^2=r^2, so y^2=r^2-x^2, so our real part equation now reads: x^3-3x(r^2-x^2) = r^3*cos(3a) = x^3-3xr^2+3x^3 = 4x^3-3xr^2 = r^3*cos(3a) or, dividing by 4, x^3-3xr^2/4-r^3*cos(3a)/4 We can now identify p = -3r^2/4, q = -r^3*cos(3a)/4 We first solve for r, r^2 = -4p/3 r = sqrt(-4p/3) So p must be negative in order to have 3 real roots. We then solve for a: cos(3a) = -4q/r^3 a = acos(-4q/r^3)/3 So |4q/r^3|<=1 in order to have 3 real roots. So our 3 real roots are: r1 = r*cos(a) r2 = r*cos(a+2pi/3) r3 = r*cos(a-2pi/3) So let us now check our work. (x-r1)(x-r2)(x-r3) = x^3 - (r1+r2+r3)x^2 + (r1r2+r1r3+r2r3)x - r1r2r3 We already know that r1+r2+r3=0, but if you want to check the trig, be my guest! -r1r2r3 = -r^3*cos(a)*cos(a+2pi/3)*cos(a-2pi/3) [after a lot of trig!] = -r^3*cos(3a)/4 r1r2+r1r3+r2r3 = r^2(cos(a)cos(a+2pi/3)+cos(a)cos(a-2pi/3)+cos(a+2pi/3)cos(a-2pi/3)) [more trig!] = r^2(-3/4) = -3r^2/4 So our equation is now x^3 - 0x^2 + (-3r^2/4)x - r^3*cos(3a)/4 = x^3-3xr^2/4-r^3*cos(3a)/4 Sure, there is a lot of trig involved in checking the solution, but since we have already motivated the solution & have it in hand, the trig involved in checking it is less stressful (although still tedious w/o Macsyma or equivalent). We do have to be careful in mapping 3 given real numbers (sum=0) to the three different angles a,a+pi/3,a-pi/3, to make sure that the mappings are consistent with the parameterization. We can now see that the case of all 3 roots being equal is handled by setting the parameter r=0, while the case of 2 roots being equal is handled by the parameter a=0 or a=pi, depending upon whether the double root is to the right or left, respectively, of the single root. An interesting fact that pops out of this characterization is a relatively tight bound, |ri|^2<r^2=-4p/3, on the size of the roots.