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x-3-x-c-0-0-lt-c-lt-2-3-3-Find-x-

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Question Number 138391 by ajfour last updated on 13/Apr/21
x^3 −x−c=0 ; 0<c<(2/(3(√3))) Find x.
x3−x−c=0;0<c<233Findx.
Answered by ajfour last updated on 13/Apr/21
let t^2 +pt+q=0 (x^3 −t^3 )−pt^2 −(x+qt)−c=0 let x−t=z (x−t){(x−t)^2 +3tx}−pt^2 −(x−t) +(q+1)t−c=0 z(z^2 +3tz+3t^2 )−z−{pt^2 −(q+1)t+c}=0 z^3 +3tz^2 +(3t^2 −1)z=pt^2 −(q+1)t+c let q=−1 pt^2 =−c & t^2 +pt−1=0 −(c/p)+pt−1=0 t=((p+c)/p^2 ) p(((p+c)/p^2 ))^2 =−c [p<0] cp^3 +p^2 +2cp+c^2 =0 p^3 +(p^2 /c)+2p+c=0 let p=s−(1/(3c)) s^3 −(s^2 /c)+(s/(3c^2 ))−(1/(27c^3 )) +(1/c)(s^2 −((2s)/(3c))+(1/(9c^2 )))+2s−(2/(3c))+c=0 ⇒ s^3 +(2−(1/(3c^2 )))s+(2/(27c^3 ))−(2/(3c))+c=0 D=((1/(27c^3 ))−(1/(3c))+(c/2))^2 −((1/(9c^2 ))−(2/3))^3 =(1/(9c^2 ))+(c^2 /4)−(2/(81c^4 ))−(1/3)+(1/(27c^2 )) +(8/(27))+(2/(81c^4 ))−(4/(27c^2 )) D=(c^2 /4)−(1/(27)) (same old tragedy..)
lett2+pt+q=0(x3−t3)−pt2−(x+qt)−c=0letx−t=z(x−t){(x−t)2+3tx}−pt2−(x−t)+(q+1)t−c=0z(z2+3tz+3t2)−z−{pt2−(q+1)t+c}=0z3+3tz2+(3t2−1)z=pt2−(q+1)t+cletq=−1pt2=−c&t2+pt−1=0−cp+pt−1=0t=p+cp2p(p+cp2)2=−c[p<0]cp3+p2+2cp+c2=0p3+p2c+2p+c=0letp=s−13cs3−s2c+s3c2−127c3+1c(s2−2s3c+19c2)+2s−23c+c=0⇒s3+(2−13c2)s+227c3−23c+c=0D=(127c3−13c+c2)2−(19c2−23)3=19c2+c24−281c4−13+127c2+827+281c4−427c2D=c24−127(sameoldtragedy..)

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