Question Number 60370 by ANTARES VY last updated on 20/May/19
Answered by math1967 last updated on 20/May/19
$$\mathrm{0} \\ $$
Answered by tanmay last updated on 20/May/19
$${N}_{{r}} =\frac{{a}^{\mathrm{2}} }{{b}+{c}}+{a}+\frac{{b}^{\mathrm{2}} }{{a}+{c}}+{b}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}+{c}−\left({a}+{b}+{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{ac}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{ab}+{bc}}{{a}+{c}}+\frac{{c}^{\mathrm{2}} +{ca}+{cb}}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\frac{{a}}{{b}+{c}}×\left({a}+{b}+{c}\right)+\frac{{b}}{{a}+{c}}×\left({a}+{b}+{c}\right)+\frac{{c}}{{a}+{b}}×\left({a}+{b}+{c}\right)−\left({a}+{b}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{a}+{c}}+\frac{{c}}{{a}+{b}}−\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$$${so}\:{given}\:{problem}\:\:=\frac{{N}_{{r}} }{{D}_{{r}} }=\frac{\mathrm{0}}{{a}+{b}+{c}}=\mathrm{0} \\ $$$$ \\ $$