Question Number 216312 by MATHEMATICSAM last updated on 03/Feb/25
$$\mathrm{If}\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\left(\frac{{a}}{{b}}\:+\:\frac{{b}}{{c}}\right)\:+\:\left(\frac{{b}}{{c}}\:+\:\frac{{c}}{{a}}\right)\:+\:\left(\frac{{c}}{{a}}\:+\:\frac{{a}}{{b}}\right)\:+\:\mathrm{2}. \\ $$
Answered by Rasheed.Sindhi last updated on 03/Feb/25
$${ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\frac{{ab}^{\mathrm{2}} }{{abc}}+\frac{{bc}^{\mathrm{2}} }{{abc}}+\frac{{ca}^{\mathrm{2}} }{{abc}}=\mathrm{0} \\ $$$$\frac{{b}}{{c}}+\frac{{c}}{{a}}+\frac{{a}}{{b}}=\mathrm{0} \\ $$$$\mathrm{2}\left(\frac{{b}}{{c}}+\frac{{c}}{{a}}+\frac{{a}}{{b}}\right)=\mathrm{0} \\ $$$$\frac{{b}}{{c}}+\frac{{c}}{{a}}+\frac{{a}}{{b}}+\frac{{b}}{{c}}+\frac{{c}}{{a}}+\frac{{a}}{{b}}=\mathrm{0} \\ $$$$\left(\frac{{a}}{{b}}+\frac{{b}}{{c}}\right)+\left(\frac{{b}}{{c}}+\frac{{c}}{{a}}\right)+\left(\frac{{c}}{{a}}+\frac{{a}}{{b}}\right)+\mathrm{2}=\mathrm{2} \\ $$
Answered by som(math1967) last updated on 03/Feb/25
$$\:\frac{\mathrm{2}{a}}{{b}}\:+\frac{\mathrm{2}{b}}{{c}}\:+\frac{\mathrm{2}{c}}{{a}}\:+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{2}} {c}+\mathrm{2}{b}^{\mathrm{2}} {a}+\mathrm{2}{c}^{\mathrm{2}} {b}}{{abc}}\:+\mathrm{2} \\ $$$$=\frac{\mathrm{2}\left({ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} \right)}{{abc}}\:+\mathrm{2} \\ $$$$=\frac{\mathrm{2}×\mathrm{0}}{{abc}}\:+\mathrm{2}=\mathrm{2} \\ $$