Rigorously over one month′s
time, I developed a formula for
general cubic.
x^3 +ax^2 +bx+c=0
let x=((pt+q)/(t+1))
pq=m, p+q=s
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m^2 {(a^2 +b)^2 −6a(ab−c)}
+m{2(b^2 +ac)(a^2 +b)−
3(ab−c)(ab+3c)}
+(b^2 +ac)^2 −6bc(ab−c)=0
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s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))}
+{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3
−8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3)
p,q = (s/2)±(√((s^2 /4)−m))
t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c)))
x=((pt+q)/(t+1)) .
(Please help checking..)
(edited a digit 1 in place of 4)
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