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Question Number 34357 by math1967 last updated on 05/May/18
If a+b+c=0  prove that  i)((a/(b+c))+ (b/(c+a)) +(c/(a+b)))(((b+c)/a) +((c+a)/b) +((a+b)/c))=9  ii)((a/(b−c)) +(b/(c−a)) +(c/(a−b)))(((b−c)/a) +((c−a)/b) +((a−b)/c))=9
$${If}\:{a}+{b}+{c}=\mathrm{0}\:\:{prove}\:{that} \\ $$$$\left.{i}\right)\left(\frac{{a}}{{b}+{c}}+\:\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}\right)\left(\frac{{b}+{c}}{{a}}\:+\frac{{c}+{a}}{{b}}\:+\frac{{a}+{b}}{{c}}\right)=\mathrm{9} \\ $$$$\left.{ii}\right)\left(\frac{{a}}{{b}−{c}}\:+\frac{{b}}{{c}−{a}}\:+\frac{{c}}{{a}−{b}}\right)\left(\frac{{b}−{c}}{{a}}\:+\frac{{c}−{a}}{{b}}\:+\frac{{a}−{b}}{{c}}\right)=\mathrm{9} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/May/18
i){((a/(b+c))+1−1)+((b/(c+a))+1−1)+((c/(a+b))+1−1)}×  {(((b+c)/a)+1−1)+(((c+a)/b)+1−1)+(((a+b)/c)+1−1)}  ={(((a+b+c)/(b+c))−1)+(((a+b+c)/(c+a))−1)+(((a+b+c)/(a+b))−1)}×  {(((a+b+c)/a)−1)+(((a+b+c)/b)−1)+(((a+b+c)/c)−1)}  =given a+b+c=0 so  =(−3)×(−3)=9
$$\left.{i}\right)\left\{\left(\frac{{a}}{{b}+{c}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{b}}{{c}+{a}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{c}}{{a}+{b}}+\mathrm{1}−\mathrm{1}\right)\right\}× \\ $$$$\left\{\left(\frac{{b}+{c}}{{a}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{c}+{a}}{{b}}+\mathrm{1}−\mathrm{1}\right)+\left(\frac{{a}+{b}}{{c}}+\mathrm{1}−\mathrm{1}\right)\right\} \\ $$$$=\left\{\left(\frac{{a}+{b}+{c}}{{b}+{c}}−\mathrm{1}\right)+\left(\frac{{a}+{b}+{c}}{{c}+{a}}−\mathrm{1}\right)+\left(\frac{{a}+{b}+{c}}{{a}+{b}}−\mathrm{1}\right)\right\}× \\ $$$$\left\{\left(\frac{{a}+{b}+{c}}{{a}}−\mathrm{1}\right)+\left(\frac{{a}+{b}+{c}}{{b}}−\mathrm{1}\right)+\left(\frac{{a}+{b}+{c}}{{c}}−\mathrm{1}\right)\right\} \\ $$$$={given}\:{a}+{b}+{c}=\mathrm{0}\:{so} \\ $$$$=\left(−\mathrm{3}\right)×\left(−\mathrm{3}\right)=\mathrm{9} \\ $$

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