Question Number 217381 by Rasheed.Sindhi last updated on 12/Mar/25
$$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{be}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{satisfying} \\ $$$$\:\mathrm{the}\:\mathrm{equations} \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{a}\:+\mathrm{b}\:+\:\mathrm{c}=\:\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} =\:\mathrm{34} \\ $$$$\mathrm{Find}\:\mathrm{ab}+\mathrm{bc}+\mathrm{ca} \\ $$
Commented by Tinku Tara last updated on 12/Mar/25
$$\mathrm{Two}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{three}\:\mathrm{variables} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solutions}. \\ $$$$ \\ $$
Commented by Tinku Tara last updated on 12/Mar/25
$$\mathrm{If}\:\mathrm{we}\:\mathrm{take}\:\mathrm{a}=\mathrm{0} \\ $$$$\mathrm{b}=\mathrm{2}−\sqrt{\frac{\mathrm{3}}{\mathrm{2}}},\:\mathrm{c}=\mathrm{2}+\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$
Commented by Tinku Tara last updated on 12/Mar/25
$${ab}+{bc}+{ca}=\mathrm{4}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{may}\:\mathrm{not}\:\mathrm{be}\:\mathrm{unique}. \\ $$
Commented by Rasheed.Sindhi last updated on 13/Mar/25
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{Tinku}\:\mathrm{Tara}\:\mathrm{sir}! \\ $$
Answered by Frix last updated on 12/Mar/25
$$\mathrm{Let}\:{a}={u}−\sqrt{{v}}\wedge{b}={u}+\sqrt{{v}} \\ $$$${a},\:{b}\:\in\mathbb{R}\:\Rightarrow\:{v}\geqslant\mathrm{0} \\ $$$$\mathrm{We}\:\:\mathrm{get} \\ $$$${v}={u}^{\mathrm{2}} −\mathrm{8}{u}+\mathrm{16}−\frac{\mathrm{5}}{{u}}\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\leqslant{u}\leqslant\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\vee{u}\geqslant\mathrm{5} \\ $$$${ab}+{bc}+{ca}=−\mathrm{4}{u}^{\mathrm{2}} +\mathrm{16}{u}−\mathrm{16}+\frac{\mathrm{5}}{{u}} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Mar/25
$$\mathbb{N}\boldsymbol{\mathrm{i}}\mathbb{C}\boldsymbol{\mathrm{e}}!\: \\ $$$$\mathcal{T}{hanks}\:{sir}! \\ $$