A few questions about solve and exact polynomial roots

_All_ cubics -- even those with complex coefficients -- can be solved with square roots, cube roots and trig functions.  Of course, you don't have to use the trig functions, but the solutions are a lot prettier & more intuitive if do you use them.

Here's a discussion for the case with 3 real roots.  Almost the same derivation works (after shifting & rotating) for the general cubic, so long as you realize that the trig functions involved now have general complex arguments.  Of course, these general trig functions can be expanded into the usual circular & hyperbolic trig functions of real arguments.  I've been preparing a similar web page with animations for the general case, but it isn't ready yet.

http://home.pipeline.com/~hbaker1/cubic3realroots.htm

At 01:27 PM 12/4/2013, Mike Valenzuela wrote:
>Wow...
>
>Thats so cool. I'm looking forward to the next version (or whichever version your function appears)!
>
>I'm not sure if there are other causes of casus irreducibilis where trig functions work better. I know that *in general* 5th order and higher order polynomials do not have closed form solutions in the field of complex radicals. I am truly clueless whether there are closed form solutions in terms of trigonometric functions. I know for a fact, that many polynomials without closed form rational solutions do have trigonometric solutions:
>x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1, has roots at 2 * cos(2*k*%pi/11), for suitable values of k
>
>I've also heard of that ultra-radicals one extention to help solve these. 
>
>I really do thank you for your work as I don't have much time as a grad-student. Unless there is a problem so ugly that I cannot do it by hand in less time than to code a functional set of rules (and I'm not very familiar with the terms I would need to make those rules in the first place), then I usually just do the ugly stuff by hand. So I want to thank you for your contribution!
>
>On Mon, Dec 2, 2013 at 3:03 PM, Aleksas Domarkas <aleksasd873 at gmail.com> wrote:
>Mike Valenzuela on  Dec 2,14:24:50 wrote:
>
>>Hello all,
>>I was messing around with some eigenvectors and values from a simple system
>>and came across the *casus irreduciblis* problem: sometimes complex
>>radicals are needed to express real solutions. However, I've been trying to
>>reduce the solution to real non-algebraic solutions.
>>(specified here)
>>http://en.wikipedia.org/wiki/Casus_irreducibilis#Non-algebraic_solution_in_terms_of_real_quantities
>>(and here)
>>http://en.wikipedia.org/wiki/Cubic_function#Trigonometric_.28and_hyperbolic.29_method
>
>>Anyways, I am wondering if maxima natively supports
>>(1) Returning non-algebraic solutions from its solving system, especially
>>if they are simplier.
>>(2) Some combination of trig simplification rules to recover the simple
>>form given in the above link.
>
>>I suppose users could manually use the above identities when applicable,
>>but they would have to be aware of it.
>
> From my "odes" package (to appear in next Maxima relase)
>function   solvet(eq,x)
>returns rectform solution of polynomial equation.
>In "casus irreducibilis" give real solutions expressed
>in trigonometric functions.
> 
> Examples:
>(%i2) solvet(x^3-3*x^2+1,x);
>(%o2) [x=2*cos(%pi/9)+1,x=2*cos((5*%pi)/9)+1,x=2*cos((7*%pi)/9)+1]
>
>(%i3) solvet(x^6-3*x^5-3*x^4+12*x^3-3*x^2-6*x+2=0,x);
>(%o3) [x=1,x=1-sqrt(3),x=sqrt(3)+1,x=2*cos((2*%pi)/9),x=2*cos((4*%pi)/9),x=2*cos((8*%pi)/9)]
>
>(%i4) solvet(x^3-15*x-5,x);
>(%o4) [x=2*sqrt(5)*cos(atan(sqrt(19))/3-(2*%pi)/3),x=2*sqrt(5)*cos(atan(sqrt(19))/3+(2*%pi)/3),x=2*sqrt(5)*cos(atan(sqrt(19))/3)]
>
>(%i5) solve(x^3-15*x-5,x);
>(%o5) [x=(-(sqrt(3)*%i)/2-1/2)*((5*sqrt(19)*%i)/2+5/2)^(1/3)+(5*((sqrt(3)*%i)/2-1/2))/((5*sqrt(19)*%i)/2+5/2)^(1/3),x=((sqrt(3)*%i)/2-1/2)*((5*sqrt(19)*%i)/2+5/2)^(1/3)+(5*(-(sqrt(3)*%i)/2-1/2))/((5*sqrt(19)*%i)/2+5/2)^(1/3),x=((5*sqrt(19)*%i)/2+5/2)^(1/3)+5/((5*sqrt(19)*%i)/2+5/2)^(1/3)]
>
>best
>
>Aleksas D