Question Number 132497 by mathocean1 last updated on 14/Feb/21
$${a},\:{b}\:\in\:\mathbb{R}. \\ $$$${Given}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −{ab}+\mathrm{11}=\mathrm{0} \\ $$$${Show}\:{that}\:−\frac{\mathrm{7}}{\mathrm{3}}<{a}+{b}<−\mathrm{2} \\ $$
Answered by MJS_new last updated on 14/Feb/21
$$\mathrm{not}\:\mathrm{true}.\:\mathrm{i}.\mathrm{e}.: \\ $$$${a}=−\mathrm{1}\wedge{b}=−\mathrm{2}\:\Rightarrow\:{a}+{b}=−\mathrm{3} \\ $$
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)](https://www.tinkutara.com/question/Q132549.png)
$${let}\:{s}={a}+{b} \\ $$$${s}^{\mathrm{3}} −\left(\mathrm{3}{s}+\mathrm{1}\right){a}\left({s}−{a}\right)+\mathrm{11}=\mathrm{0} \\ $$$$\left(\mathrm{3}{s}+\mathrm{1}\right){a}^{\mathrm{2}} −{s}\left(\mathrm{3}{s}+\mathrm{1}\right){a}+{s}^{\mathrm{3}} +\mathrm{11}=\mathrm{0} \\ $$$$\Delta={s}^{\mathrm{2}} \left(\mathrm{3}{s}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}{s}+\mathrm{1}\right)\left({s}^{\mathrm{3}} +\mathrm{11}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{s}+\mathrm{1}\right)\left({s}^{\mathrm{2}} −{s}^{\mathrm{3}} −\mathrm{44}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{s}+\mathrm{1}\right)\left({s}^{\mathrm{3}} −{s}^{\mathrm{2}} +\mathrm{44}\right)\leqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{3}{s}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{s}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${s}^{\mathrm{3}} −{s}^{\mathrm{2}} +\mathrm{44}=\mathrm{0} \\ $$$${let}\:{s}={t}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{{t}}{\mathrm{3}}+\frac{\mathrm{1186}}{\mathrm{27}}=\mathrm{0} \\ $$$${t}=−\sqrt[{\mathrm{3}}]{\frac{\mathrm{593}}{\mathrm{27}}+\sqrt{\frac{\mathrm{593}^{\mathrm{2}} }{\mathrm{27}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} }}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{593}}{\mathrm{27}}−\sqrt{\frac{\mathrm{593}^{\mathrm{2}} }{\mathrm{27}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} }}} \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{593}+\mathrm{12}\sqrt{\mathrm{2442}}}−\sqrt[{\mathrm{3}}]{\mathrm{593}−\mathrm{12}\sqrt{\mathrm{2442}}}\right]\approx−\mathrm{3}.\mathrm{2265} \\ $$$$ \\ $$$$\Rightarrow−\mathrm{3}.\mathrm{2265}\leqslant{s}={a}+{b}\leqslant−\frac{\mathrm{1}}{\mathrm{3}} \\ $$