Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 35080 by math1967 last updated on 15/May/18

If (a/(b+c)) +(b/(c+a)) +(c/(a+b))=1 then prove that (a^2 /(b+c)) +(b^2 /(c+a)) +(c^2 /(a+b))=0

$${If}\:\frac{{a}}{{b}+{c}}\:+\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}=\mathrm{1}\:{then}\:{prove}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+\frac{{c}^{\mathrm{2}} }{{a}+{b}}=\mathrm{0} \\ $$

Answered by ajfour last updated on 15/May/18

(a/(b+c))+(b/(c+a))+(c/(a+b))=1 (a+b+c)[(a/(b+c))+(b/(c+a))+(c/(a+b))]=a+b+c .......(i) now let (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b))= q (i) gives q+((a(b+c))/(b+c))+((b(c+a))/(c+a))+((c(a+b))/(a+b))=a+b+c q= (a^2 /(b+c))+(b^2 /(c+a))+(c^2 /(a+b)) = 0 .

$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}=\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)\left[\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}\right]={a}+{b}+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......\left({i}\right) \\ $$$${now}\:{let}\:\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}=\:{q} \\ $$$$\left({i}\right)\:{gives} \\ $$$${q}+\frac{{a}\left({b}+{c}\right)}{{b}+{c}}+\frac{{b}\left({c}+{a}\right)}{{c}+{a}}+\frac{{c}\left({a}+{b}\right)}{{a}+{b}}={a}+{b}+{c} \\ $$$$\:\:\:{q}=\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:=\:\mathrm{0}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18

=(a^2 /(b+c))+a+(b^2 /(c+a))+b+(c^2 /(a+b))+c−(a+b+c) ((a^2 +ab+ac)/(b+c))+((b^2 +bc+ab)/(c+a))+((c^2 +ac+bc)/(a+b))−(a+b+c) =((a(a+b+c))/(b+c))+((b(a+b+c))/(c+a))+((c(a+b+c))/(a+b))−(a+b+c) =(a+b+c)((a/(b+c))+(b/(c+a))+(c/(a+b))−1) =(a+b+c)(1−1) =0

$$=\frac{{a}^{\mathrm{2}} }{{b}+{c}}+{a}+\frac{{b}^{\mathrm{2}} }{{c}+{a}}+{b}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}+{c}−\left({a}+{b}+{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{ac}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{bc}+{ab}}{{c}+{a}}+\frac{{c}^{\mathrm{2}} +{ac}+{bc}}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\frac{{a}\left({a}+{b}+{c}\right)}{{b}+{c}}+\frac{{b}\left({a}+{b}+{c}\right)}{{c}+{a}}+\frac{{c}\left({a}+{b}+{c}\right)}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}−\mathrm{1}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\mathrm{1}−\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com