如何解3次方程式？ax^3+bx^2+cx+d=0
exact solution
consider x^3+ax^2+b*x+c=0 (1)
omega = (-1+3^(1/2)*i)/2
x = y-a/3 substitute into (1) obtain y^3+py+q=0
and let
s=(-q/2+((q/2)^2+(p/3)^3)^(1/2))^(1/3)
t=(-q/2-((q/2)^2+(p/3)^3)^(1/2))^(1/3)
then
y1 = s+t
y2 = s*omega+t*omega^2
y3 = s*omega^2+t*omega
omega = (-1+3^(1/2)*i)/2
x = y-a/3 substitute into (1) obtain y^3+py+q=0
and let
s=(-q/2+((q/2)^2+(p/3)^3)^(1/2))^(1/3)
t=(-q/2-((q/2)^2+(p/3)^3)^(1/2))^(1/3)
then
y1 = s+t
y2 = s*omega+t*omega^2
y3 = s*omega^2+t*omega
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