x^3 +ax^2 +bx+c=0
Let x=((pt+q)/(t+1))
p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3
+a(t+1)(p^2 t^2 +2pqt+q^2 )
+b(pt+q)(t^2 +2t+1)
+c(t^3 +3t^2 +3t+1) = 0
⇒
(p^3 +ap^2 +bp+c)t^3
+(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2
+(3q^2 p+aq^2 +2apq+bp+2bq+3c)t
+(q^3 +aq^2 +bq+c) = 0
Let coeffs. of t^2 and t be zero.
Subtracting and adding them
3pq+a(p+q)+b=0 &
3pq(p+q)+a{(p+q)^2 −2pq}
+4apq+3b(p+q)+6c = 0
lets call pq=m , p+q=s ⇒
3m+as+b=0 ....(i)
3ms+a(s^2 −2m)+4am+3bs+6c=0
⇒ am+bs+3c=0 ....(ii)
⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b))
Now p,q are roots of eq.
z^2 −sz+m=0
p,q = (s/2)±(√((s^2 /4)−m))
t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1)
x=((pt+q)/(t+1)) .
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