Hi,
I have polynomial equation :
b3:c^3+2*c^2+(1-P)*c+(P-1)^2=0$
where c is variable
I can solve it in Maxima :
======= code ===============
(%i3) b3:c^3+2*c^2+(1-P)*c+(P-1)^2=0$
(%i4) solve(b3,c);
(%o4)
[c=(-(sqrt(3)*%i)/2-1/2)*(((P-1)*sqrt(27*P^2-22*P+23))/(6*sqrt(3))-(27*P^2-36*P+25)/54)^(1/3)+
(((sqrt(3)*%i)/2-1/2)*(3*P+1))/(9*(((P-
1)*sqrt(27*P^2-22*P+23))/(6*sqrt(3))-(27*P^2-36*P+25)/54)^(1/3))-2/3,c=((sqrt(3)*%i)/2-1/2)*
(((P-1)*sqrt(27*P^2-22*P+23))/(6*sqrt(3))-(27*P^2-36*P+25)/54)^(1/3)+
((-(sqrt(3)*%i)/2-1/2)*(3*P+1))/(9*(((P-1)*sqrt(27*P^2-22*P+23))
/(6*sqrt(3))-
(27*P^2-36*P+25)/54)^(1/3))-2/3,c=(((P-1)*sqrt(27*P^2-22*P+23))/(6*sqrt(3))-(27*P^2-36*P+25)/54)^(1/3)+
(3*P+1)/(9*(((P-1)*sqrt(27*P^2-22*P+23))/(6*sqrt(3))-(27*P^2-36*P+25)/54)^(1/3))
============================
It gives 3 functions. Each functions maps one half of one component and
one half of other component.
http://commons.wikimedia.org/wiki/Image:HypComp3.png
Why it is ?
It is possible to have other solution : one function for one component
Regards
Adam