Question Number 187858 by normans last updated on 23/Feb/23
Answered by som(math1967) last updated on 23/Feb/23
$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−{abc}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}+{abc}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{abc}+\cancel{{abc}} \\ $$$$\:+{bc}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}−\cancel{{abc}}=\mathrm{0} \\ $$$$\Rightarrow{ab}\left({a}+{c}\right)+{b}^{\mathrm{2}} \left({a}+{c}\right)+{ca}\left({a}+{c}\right)+{bc}\left({a}+{c}\right)=\mathrm{0} \\ $$$$\left({a}+{c}\right)\left({ab}+{b}^{\mathrm{2}} +{ca}+{bc}\right)=\mathrm{0} \\ $$$$\left({a}+{c}\right)\left\{{b}\left({a}+{b}\right)+{c}\left({a}+{b}\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{c}\right)\left({a}+{b}\right)\left({b}+{c}\right)=\mathrm{0} \\ $$$${if}\:\left({a}+{c}\right)=\mathrm{0}\therefore{a}=−{c} \\ $$$$\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=−\frac{\mathrm{1}}{{c}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=\frac{\mathrm{1}}{{b}^{\mathrm{5}} } \\ $$$${again}\:\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{5}} }=\frac{\mathrm{1}}{{b}^{\mathrm{5}} } \\ $$$${if}\:{a}+{b}=\mathrm{0}\Rightarrow{a}=−{b}\:{orb}=−{c} \\ $$$${gives}\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{5}} } \\ $$$$ \\ $$
Commented by normans last updated on 23/Feb/23
$${very}\:{nice}\:{your}\:{solution}\:{Sir} \\ $$