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a-b-R-Given-a-3-b-3-ab-11-0-Show-that-7-3-lt-a-b-lt-2-

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Question Number 132497 by mathocean1 last updated on 14/Feb/21
a, b ∈ R. Given a^3 +b^3 −ab+11=0 Show that −(7/3)<a+b<−2
a,b∈R.Givena3+b3−ab+11=0Showthat−73<a+b<−2
Answered by MJS_new last updated on 14/Feb/21
not true. i.e.: a=−1∧b=−2 ⇒ a+b=−3
nottrue.i.e.:a=−1∧b=−2⇒a+b=−3
Answered by mr W last updated on 15/Feb/21
let s=a+b s^3 −(3s+1)a(s−a)+11=0 (3s+1)a^2 −s(3s+1)a+s^3 +11=0 Δ=s^2 (3s+1)^2 −4(3s+1)(s^3 +11)≥0 (3s+1)(s^2 −s^3 −44)≥0 (3s+1)(s^3 −s^2 +44)≤0 3s+1=0 ⇒s=−(1/3) s^3 −s^2 +44=0 let s=t+(1/3) t^3 −(t/3)+((1186)/(27))=0 t=−((((593)/(27))+(√(((593^2 )/(27^2 ))−(1/9^3 )))))^(1/3) −((((593)/(27))−(√(((593^2 )/(27^2 ))−(1/9^3 )))))^(1/3) s=(1/3)[1−((593+12(√(2442))))^(1/3) −((593−12(√(2442))))^(1/3) ]≈−3.2265 ⇒−3.2265≤s=a+b≤−(1/3)
lets=a+bs3−(3s+1)a(s−a)+11=0(3s+1)a2−s(3s+1)a+s3+11=0Δ=s2(3s+1)2−4(3s+1)(s3+11)⩾0(3s+1)(s2−s3−44)⩾0(3s+1)(s3−s2+44)⩽03s+1=0⇒s=−13s3−s2+44=0lets=t+13t3−t3+118627=0t=−59327+5932272−1933−59327−5932272−1933s=13[1−593+1224423−593−1224423]≈−3.2265⇒−3.2265⩽s=a+b⩽−13

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