Question Number 191615 by MATHEMATICSAM last updated on 27/Apr/23
$${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that}, \\ $$$$\frac{{a}}{{a}^{\mathrm{2}} \:−\:{bc}}\:+\:\frac{{b}}{{b}^{\mathrm{2}} \:−\:{ca}}\:+\:\frac{{c}}{{c}^{\mathrm{2}} \:−\:{ab}}\:=\:\mathrm{0}. \\ $$
Answered by som(math1967) last updated on 27/Apr/23
$$\:{a}+{b}+{c}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =−{ab}−{ca} \\ $$$${b}^{\mathrm{2}} =−{bc}−{ab},\:{c}^{\mathrm{2}} =−{ca}−{bc} \\ $$$$\frac{{a}}{{a}^{\mathrm{2}} −{bc}}+\frac{{b}}{{b}^{\mathrm{2}} −{ca}}+\frac{{c}}{{c}^{\mathrm{2}} −{ab}} \\ $$$$=\frac{{a}}{−\left({ab}+{bc}+{ca}\right)}+\frac{{b}}{−\left({ab}+{bc}+{ca}\right)} \\ $$$$\:\:+\frac{{c}}{−\left({ab}+{bc}+{ca}\right)} \\ $$$$=\frac{{a}+{b}+{c}}{−\left({ab}+{bc}+{ca}\right)}=\frac{\mathrm{0}}{−\left({ab}+{bc}+{ca}\right)}=\mathrm{0} \\ $$