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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=0 Showthat−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+b s3−(3s+1)a(s−a)+11=0 (3s+1)a2−s(3s+1)a+s3+11=0 Δ=s2(3s+1)2−4(3s+1)(s3+11)⩾0 (3s+1)(s2−s3−44)⩾0 (3s+1)(s3−s2+44)⩽0 3s+1=0 ⇒s=−13 s3−s2+44=0 lets=t+13 t3−t3+118627=0 t=−59327+5932272−1933−59327−5932272−1933 s=13[1−593+1224423−593−1224423]≈−3.2265 ⇒−3.2265⩽s=a+b⩽−13

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