Question Number 194971 by kapoorshah last updated on 21/Jul/23
Commented by Frix last updated on 21/Jul/23
$$\mathrm{I}\:\mathrm{think}\:\left({abc}\right)^{\mathrm{2}} =\mathrm{8} \\ $$
Commented by sniper237 last updated on 23/Jul/23
$${a}+\frac{\mathrm{2}}{{b}}={b}+\frac{\mathrm{2}}{{c}}\:\Rightarrow\:{a}\:−{b}\:=\:\frac{\mathrm{2}}{{c}}\:−\frac{\mathrm{2}}{{b}}\:\Rightarrow\:{bc}\:=\:\frac{\mathrm{2}\left({b}−{c}\right)}{{a}−{b}} \\ $$$${Samely}\:\:{ac}\:=\:\frac{\mathrm{2}\left({c}−{a}\right)}{{b}−{c}}\:{and}\:\:{ab}\:=\:\frac{\mathrm{2}\left({b}−{a}\right)}{{a}−{c}} \\ $$$${timing}\:\:{all}\:\:\left({abc}\right)^{\mathrm{2}} =\left({ab}\right)\left({ac}\right)\left({bc}\right)=\mathrm{8} \\ $$
Answered by Frix last updated on 21/Jul/23
$${a}+\frac{\mathrm{2}}{{b}}={b}+\frac{\mathrm{2}}{{c}}\wedge{a}+\frac{\mathrm{2}}{{b}}={c}+\frac{\mathrm{2}}{{a}} \\ $$$${c}=\frac{\mathrm{2}{b}}{{ab}−{b}^{\mathrm{2}} +\mathrm{2}}=\frac{{a}^{\mathrm{2}} {b}+\mathrm{2}{a}−\mathrm{2}{b}}{{ab}} \\ $$$$\mathrm{Transforming}\:\mathrm{and}\:\mathrm{factorizing}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\left({a}−{b}\right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{4}{ab}−\mathrm{2}{b}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{But}\:{a}\neq{b}\:\Rightarrow \\ $$$${b}=−\frac{\mathrm{2}\left({a}\pm\sqrt{\mathrm{2}}\right)}{{a}^{\mathrm{2}} −\mathrm{2}^{\left(\ast\right)} }\wedge{c}=−\frac{\mathrm{2}\mp{a}\sqrt{\mathrm{2}}}{{a}} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{8} \\ $$$$\:^{\left(\ast\right)} \:\mathrm{Set}\:{a}=\sqrt{\mathrm{2}}\:\mathrm{from}\:\mathrm{the}\:\mathrm{beginning}\:\Rightarrow\:\mathrm{we}\:\mathrm{get} \\ $$$$\:\:\:\:\:\:\:{b}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge{c}=−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{still}\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{8} \\ $$
Commented by kapoorshah last updated on 21/Jul/23
$${nice} \\ $$$${thank}\:{you} \\ $$