Question Number 102490 by ajfour last updated on 09/Jul/20

x^3 −bx−c=0     ;  b, c >0 ;  ((b/3))^3 >((c/2))^2   To find the three real roots without  the use of trigonometric solution  to cubic polynomial...

Answered by ajfour last updated on 10/Jul/20

let  z=(x/(√b)) , a=(c/(b(√b)))    As   (c/2)<((b(√b))/(3(√3)))   ⇒   a<(2/(3(√3)))  ⇒  z^3 −z−a=0  Now let    (z−1)(z^3 −z−a)=0      z^4 −z^3 −z^2 +(1−a)z+a=0  Now let z=((t+s)/(t+1))   ⇒   (t^4 +4st^3 +6s^2 t^2 +4s^3 t+s^4 )  −(t+1)(t^3 +3st^2 +3s^2 t+s^3 )  −(t^2 +2t+1)(t^2 +2st+s^2 )  +(1−a)(t+s)(t^3 +3t^2 +3t+1)  +a(t^4 +4t^3 +6t^2 +4t+1)=0  ⇒  t^3 {4s−3s−1−2s−2+3(1−a)         +s(1−a)+4a}+  t^2 {6s^2 −3s^2 −3s−s^2 −4s−1        +3(1−a)+3s(1−a)+6a}+  t{4s^3 −s^3 −3s^2 −2s^2 −2s       +(1−a)+3s(1−a)+4a}+     {s^4 −s^3 −s^2 +s(1−a)+a} = 0  ⇒     t^3 {a(1−s)}+t^2 {2s^2 −s(4+3a)+3a+2}  +t{s^3 −5s^2 −(3a−1)s+3a+1}  +{s^4 −s^3 −s^2 +s(1−a)+a}=0  Now let   2s^2 −s(4+3a)+3a+2=0  ⇒   s=((3a+4)/2)±((√((3a+4)^2 −2(3a+2)))/2)            = ((3a+4±(√((3a+4)^2 −2(3a+4)+4)))/2)            =((3(a+1)+1±(√(9(a+1)^2 +3)))/2)     Now         t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))}         +{((s^4 −s^3 −s^2 +s(1−a)+a)/(a(1−s)))}= 0  ⇒  t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))}             −(1/a)(s^3 −s−a) = 0  ⇒  t^3 +(t/a){(s+3)^2 +3a+4(((2−s)/(s−1)))}             −(1/a)(s^3 −s−a) = 0  lets say above expression is            t^3 +((p/a))t−(q/a)=0  we can apply Cardano′s formula if  p>0 ;  say t=t_0   , then              z_0 =((t_0 +s)/(t_0 +1))    and  x=(√b)(z_0 )  .......  lets check the sign of     p = (s+3)^2 +3a+4(((2−s)/(s−1)))    where           s =((3(a+1)+1±(√(9(a+1)^2 +3)))/2)          and a<(2/(3(√3)))  (someone please help here...      can p be >0 for one of the two      values of s ? )