| z^3 βz=c
z^4 βz^2 =cz
let z=x+q
x^4 +4qx^3 +6q^2 x^2 +4q^3 x+q^4
βx^2 β2qxβq^2 βcxβcq=0
β
x^4 +4qx^3 +(6q^2 β1)x^2 +
(4q^3 β2qβc)x+q^4 βq^2 βcq=0
let q^2 =1/6 β
x^4 +4qx^3 β(((4q)/3)+c)xβ(5/(36))βcq=0
β‘ (x^2 +mx+k)(x^2 +sx+h)=0
β m+s=4q
h+k+ms=0
mh+ks=β(((4q)/3)+c)
kh=β(5/(36))βcq
h, k= β((ms)/2)Β±(β(((m^2 s^2 )/4)+(5/(36))+cq))
m(β((ms)/2)+D)+s(β((ms)/2)βD)
+((4q)/3)+c=0
(mβs)Dβ((ms)/2)(4q)+((4q)/3)+c=0
now letβ²s assume ms=p
& since m+s=4q
(mβs)^2 =(8/3)β4p
β ((8/3)β4p)((p^2 /4)+(5/(36))+cq)
=(2pqβ((4q)/3)βc)^2
p^3 +((5/9)+4cqβ(8/9)β4cq)p
+(((4q)/3)+c)^2 β(8/3)((5/(36))+cq)=0
β p^3 β(p/3)+((14)/(27))=0
D_0 =((7/(27)))^2 β((1/9))^3
....
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