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Question Number 8355 by Nayon last updated on 09/Oct/16
if a+b+c=0 then (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b)))=?
$${if}\:{a}+{b}+{c}=\mathrm{0}\:{then} \\ $$$$ \\ $$$$\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right)=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 11/Oct/16
=(((a−b)/c))((a/(b−c))+(b/(c−a))+(c/(a−b))) +(((b−c)/a))((a/(b−c))+(b/(c−a))+(c/(a−b))) +(((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b))) ={(((a−b)/c))((a/(b−c))+(b/(c−a)))+1} +{(((b−c)/a))((b/(c−a))+(c/(a−b)))+1} +{(((c−a)/b))((a/(b−c))+(c/(a−b)))+1} =(((a−b)/c))((a/(b−c))+(b/(c−a))) +(((b−c)/a))((b/(c−a))+(c/(a−b))) +(((c−a)/b))((a/(b−c))+(c/(a−b))) +3 =(((a−b)/c))(((ac−a^2 +b^2 −bc)/((b−c)(c−a)))) +(((b−c)/a))(((ab−b^2 +c^2 −ac)/((c−a)(a−b)))) +(((c−a)/b))(((a^2 −ab+bc−c^2 )/((b−c)(a−b)))) +3 =(((a−b)/c))(((−(a−b)(a+b)+c(a−b))/((b−c)(c−a)))) +(((b−c)/a))(((−(b−c)(b+c)+a(b−c))/((c−a)(a−b)))) +(((c−a)/b))(((−(c−a)(c+a)−b(c−a))/((b−c)(a−b)))) +3 =(((a−b)/c))((((a−b)(−a−b+c))/((b−c)(c−a)))) +(((b−c)/a))((((b−c)(−b−c+a))/((c−a)(a−b)))) +(((c−a)/b))((((c−a){−(a+b+c)})/((b−c)(a−b)))) +3 =(((a−b)^2 (−a−b+c))/(c(b−c)(c−a))) +(((b−c)^2 (−b−c+a))/(a(c−a)(a−b))) +(((c−a)^2 {−(0)})/(b(b−c)(a−b))) +3 =(((a−b)^2 (−a−b+c))/(c(b−c)(c−a)))+(((b−c)^2 (a−b−c))/(a(c−a)(a−b)))+3 =((1/(c−a)))((((a−b)^2 (−a−b+c))/(c(b−c)))+(((b−c)^2 (a−b−c))/(a(a−b))))+3 =((1/(c−a)))((((a−b)^2 (−a−b+c)+(b−c)^2 (a−b−c))/(ac(a−b)(b−c))))+3 =((1/(c−a)))((((a−b)^2 (a+b+c−2a−2b)+(b−c)^2 (a+b+c−2b−2c))/(ac(a−b)(b−c))))+3 =((1/(c−a)))((((a−b)^2 (−2a−2b)+(b−c)^2 (−2b−2c))/(ac(a−b)(b−c))))+3 =(((−2)/(c−a)))((((a−b)^2 (a+b)+(b−c)^2 (b+c))/(ac(a−b)(b−c))))+3 Continue to the next comment.
$$=\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$=\left\{\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}\right)+\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left\{\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right)+\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left\{\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{c}}{{a}−{b}}\right)+\mathrm{1}\right\} \\ $$$$=\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$=\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{{ac}−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{bc}}{\left({b}−{c}\right)\left({c}−{a}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{{ab}−{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ac}}{\left({c}−{a}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}^{\mathrm{2}} −{ab}+{bc}−{c}^{\mathrm{2}} }{\left({b}−{c}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$=\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{−\left({a}−{b}\right)\left({a}+{b}\right)+{c}\left({a}−{b}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{−\left({b}−{c}\right)\left({b}+{c}\right)+{a}\left({b}−{c}\right)}{\left({c}−{a}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{−\left({c}−{a}\right)\left({c}+{a}\right)−{b}\left({c}−{a}\right)}{\left({b}−{c}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$=\left(\frac{{a}−{b}}{{c}}\right)\left(\frac{\left({a}−{b}\right)\left(−{a}−{b}+{c}\right)}{\left({b}−{c}\right)\left({c}−{a}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{b}−{c}}{{a}}\right)\left(\frac{\left({b}−{c}\right)\left(−{b}−{c}+{a}\right)}{\left({c}−{a}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}−{a}}{{b}}\right)\left(\frac{\left({c}−{a}\right)\left\{−\left({a}+{b}+{c}\right)\right\}}{\left({b}−{c}\right)\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$=\frac{\left({a}−{b}\right)^{\mathrm{2}} \left(−{a}−{b}+{c}\right)}{\mathrm{c}\left({b}−{c}\right)\left({c}−{a}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\left({b}−{c}\right)^{\mathrm{2}} \left(−{b}−{c}+{a}\right)}{{a}\left({c}−{a}\right)\left({a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\left({c}−{a}\right)^{\mathrm{2}} \left\{−\left(\mathrm{0}\right)\right\}}{{b}\left({b}−{c}\right)\left({a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$=\frac{\left({a}−{b}\right)^{\mathrm{2}} \left(−{a}−{b}+{c}\right)}{\mathrm{c}\left({b}−{c}\right)\left({c}−{a}\right)}+\frac{\left({b}−{c}\right)^{\mathrm{2}} \left({a}−{b}−{c}\right)}{{a}\left({c}−{a}\right)\left({a}−{b}\right)}+\mathrm{3} \\ $$$$=\left(\frac{\mathrm{1}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left(−{a}−{b}+{c}\right)}{\mathrm{c}\left({b}−{c}\right)}+\frac{\left({b}−{c}\right)^{\mathrm{2}} \left({a}−{b}−{c}\right)}{{a}\left({a}−{b}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{\mathrm{1}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left(−{a}−{b}+{c}\right)+\left({b}−{c}\right)^{\mathrm{2}} \left({a}−{b}−{c}\right)}{{a}\mathrm{c}\left({a}−{b}\right)\left({b}−{c}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{\mathrm{1}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}+{c}−\mathrm{2}{a}−\mathrm{2}{b}\right)+\left({b}−{c}\right)^{\mathrm{2}} \left({a}+{b}+{c}−\mathrm{2}{b}−\mathrm{2}{c}\right)}{{a}\mathrm{c}\left({a}−{b}\right)\left({b}−{c}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{\mathrm{1}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left(−\mathrm{2}{a}−\mathrm{2}{b}\right)+\left({b}−{c}\right)^{\mathrm{2}} \left(−\mathrm{2}{b}−\mathrm{2}{c}\right)}{{a}\mathrm{c}\left({a}−{b}\right)\left({b}−{c}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{−\mathrm{2}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)+\left({b}−{c}\right)^{\mathrm{2}} \left({b}+{c}\right)}{{a}\mathrm{c}\left({a}−{b}\right)\left({b}−{c}\right)}\right)+\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Continue}\:\mathrm{to}\:\mathrm{the}\:\mathrm{next}\:\mathrm{comment}. \\ $$
Commented by Rasheed Soomro last updated on 13/Oct/16
Continue from above a+b+c=0⇒c=−a−b =(((−2)/(c−a)))((((a−b)^2 (a+b)+(b−c)^2 (b+c))/(ac(a−b)(b−c))))+3 =(((−2)/(−a−b−a)))((((a−b)^2 (a+b)+(b−{−a−b})^2 (−a))/(a(−a−b)(a−b)(b−{−a−b}))))+3 =(((−2)/(−2a−b)))((((a−b)^2 (a+b)+(2b+a)^2 (−a))/(a(−a−b)(a−b)(2b+a))))+3 =−((2/(2a+b)))((((a−b)^2 (a+b)−a(2b+a)^2 )/(a(a+b)(a−b)(2b+a))))+3 =−2((((a−b)^2 (a+b)−a(2b+a)^2 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2((((a^2 −b^2 )(a−b)−a(4b^2 +4ab+a^2 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2(((−ab^2 −a^2 b+b^3 −4ab^2 −4a^2 b)/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2(((−5ab^2 −5a^2 b+b^3 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3
$$\mathrm{Continue}\:\mathrm{from}\:\mathrm{above} \\ $$$${a}+{b}+{c}=\mathrm{0}\Rightarrow\mathrm{c}=−{a}−{b} \\ $$$$=\left(\frac{−\mathrm{2}}{{c}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)+\left({b}−{c}\right)^{\mathrm{2}} \left({b}+{c}\right)}{{a}\mathrm{c}\left({a}−{b}\right)\left({b}−{c}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{−\mathrm{2}}{−{a}−{b}−{a}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)+\left({b}−\left\{−{a}−{b}\right\}\right)^{\mathrm{2}} \left(−{a}\right)}{{a}\left(−{a}−{b}\right)\left({a}−{b}\right)\left({b}−\left\{−{a}−{b}\right\}\right)}\right)+\mathrm{3} \\ $$$$=\left(\frac{−\mathrm{2}}{−\mathrm{2}{a}−{b}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)+\left(\mathrm{2}{b}+{a}\right)^{\mathrm{2}} \left(−{a}\right)}{{a}\left(−{a}−{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)}\right)+\mathrm{3} \\ $$$$=−\left(\frac{\mathrm{2}}{\mathrm{2}{a}+{b}}\right)\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)−{a}\left(\mathrm{2}{b}+{a}\right)^{\mathrm{2}} }{{a}\left({a}+{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)}\right)+\mathrm{3} \\ $$$$=−\mathrm{2}\left(\frac{\left({a}−{b}\right)^{\mathrm{2}} \left({a}+{b}\right)−{a}\left(\mathrm{2}{b}+{a}\right)^{\mathrm{2}} }{{a}\left({a}+{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)\left(\mathrm{2}{a}+{b}\right)}\right)+\mathrm{3} \\ $$$$=−\mathrm{2}\left(\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({a}−{b}\right)−{a}\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{ab}+{a}^{\mathrm{2}} \right.}{{a}\left({a}+{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)\left(\mathrm{2}{a}+{b}\right)}\right)+\mathrm{3} \\ $$$$=−\mathrm{2}\left(\frac{−{ab}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} −\mathrm{4}{ab}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {b}}{{a}\left({a}+{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)\left(\mathrm{2}{a}+{b}\right)}\right)+\mathrm{3} \\ $$$$=−\mathrm{2}\left(\frac{−\mathrm{5}{ab}^{\mathrm{2}} −\mathrm{5}{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} }{{a}\left({a}+{b}\right)\left({a}−{b}\right)\left(\mathrm{2}{b}+{a}\right)\left(\mathrm{2}{a}+{b}\right)}\right)+\mathrm{3} \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 12/Oct/16
a+b+c=0 , (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b)))=? −−−−−−−−−−−−−−−−−−−−−−−−−−−− a+b+c=0⇒c=−a−b (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b))) =(((a−b)/(−a−b))+((b−(−a−b))/a)+((−a−b−a)/b))((a/(b−(−a−b)))+(b/(−a−b−a))+((−a−b)/(a−b))) =(− ((a−b)/(a+b))+((a+2b)/a)− ((2a+b)/b))((a/(a+2b))− (b/(2a+b))− ((a+b)/(a−b))) =(− ((a−b)/(a+b)))((a/(a+2b))− (b/(2a+b))− ((a+b)/(a−b))) +(((a+2b)/a))((a/(a+2b))− (b/(2a+b))− ((a+b)/(a−b))) +(− ((2a+b)/b))((a/(a+2b))− (b/(2a+b))− ((a+b)/(a−b))) =(−((a(a−b))/((a+b)(a+2b)))+((b(a−b))/((a+b)(2a+b)))+1) +(1−((b(a+2b))/(a(2a+b)))−(((a+b)(a+2b))/(a(a−b)))) +(− ((a(2a+b))/(b(a+2b))+1+(((a+b)(2a+b))/(b(a−b)))) =−((a(a−b))/((a+b)(a+2b)))+((b(a−b))/((a+b)(2a+b))) − ((b(a+2b))/(a(2a+b)))−(((a+b)(a+2b))/(a(a−b))) − ((a(2a+b))/(b(a+2b))+(((a+b)(2a+b))/(b(a−b))) +3 = Continue
$${a}+{b}+{c}=\mathrm{0}\:,\:\:\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right)=? \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${a}+{b}+{c}=\mathrm{0}\Rightarrow{c}=−{a}−{b} \\ $$$$\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\frac{{a}−{b}}{−{a}−{b}}+\frac{{b}−\left(−{a}−{b}\right)}{{a}}+\frac{−{a}−{b}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−\left(−{a}−{b}\right)}+\frac{{b}}{−{a}−{b}−{a}}+\frac{−{a}−{b}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left(−\:\frac{{a}−{b}}{{a}+{b}}+\frac{{a}+\mathrm{2}{b}}{{a}}−\:\frac{\mathrm{2}{a}+{b}}{{b}}\right)\left(\frac{{a}}{{a}+\mathrm{2}{b}}−\:\frac{{b}}{\mathrm{2}{a}+{b}}−\:\frac{{a}+{b}}{{a}−{b}}\right) \\ $$$$=\left(−\:\frac{{a}−{b}}{{a}+{b}}\right)\left(\frac{{a}}{{a}+\mathrm{2}{b}}−\:\frac{{b}}{\mathrm{2}{a}+{b}}−\:\frac{{a}+{b}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{a}+\mathrm{2}{b}}{{a}}\right)\left(\frac{{a}}{{a}+\mathrm{2}{b}}−\:\frac{{b}}{\mathrm{2}{a}+{b}}−\:\frac{{a}+{b}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(−\:\frac{\mathrm{2}{a}+{b}}{{b}}\right)\left(\frac{{a}}{{a}+\mathrm{2}{b}}−\:\frac{{b}}{\mathrm{2}{a}+{b}}−\:\frac{{a}+{b}}{{a}−{b}}\right) \\ $$$$=\left(−\frac{{a}\left({a}−{b}\right)}{\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)}+\frac{{b}\left({a}−{b}\right)}{\left({a}+{b}\right)\left(\mathrm{2}{a}+{b}\right)}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{1}−\frac{{b}\left({a}+\mathrm{2}{b}\right)}{{a}\left(\mathrm{2}{a}+{b}\right)}−\frac{\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)}{{a}\left({a}−{b}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(−\:\frac{{a}\left(\mathrm{2}{a}+{b}\right)}{{b}\left({a}+\mathrm{2}{b}\right.}+\mathrm{1}+\frac{\left({a}+{b}\right)\left(\mathrm{2}{a}+{b}\right)}{{b}\left({a}−{b}\right)}\right) \\ $$$$=−\frac{{a}\left({a}−{b}\right)}{\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)}+\frac{{b}\left({a}−{b}\right)}{\left({a}+{b}\right)\left(\mathrm{2}{a}+{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\frac{{b}\left({a}+\mathrm{2}{b}\right)}{{a}\left(\mathrm{2}{a}+{b}\right)}−\frac{\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)}{{a}\left({a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\frac{{a}\left(\mathrm{2}{a}+{b}\right)}{{b}\left({a}+\mathrm{2}{b}\right.}+\frac{\left({a}+{b}\right)\left(\mathrm{2}{a}+{b}\right)}{{b}\left({a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3} \\ $$$$= \\ $$$$ \\ $$$${C}\mathrm{ontinue} \\ $$
Answered by Rasheed Soomro last updated on 11/Oct/16
a+b+c=0 , (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b)))=? (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b))) =(((a+b+c−2b−c)/c)+((a+b+c−a−2c)/a)+((a+b+c−2a−b)/b)) ×((a/(a+b+c−a−2c))+(b/(a+b+c−2a−b))+(c/(a+b+c−2b−c))) =(((−2b−c)/c)+((−a−2c)/a)+((−2a−b)/b)) ×((a/(−a−2c))+(b/(−2a−b))+(c/(−2b−c))) a+b+c=0⇒c=−a−b =(((−2b−(−a−b))/(−a−b))+((−a−2(−a−b))/a)+((−2a−b)/b)) ×((a/(−a−2(−a−b)))+(b/(−2a−b))+((−a−b)/(−2b−(−a−b))) =(((−2b+a+b))/(−a−b))+((−a+2a+2b))/a)+((−2a−b)/b)) ×((a/(−a+2a+2b)))+(b/(−2a−b))+((−a−b)/(−2b+a+b))) =(((a−b)/(−a−b))+1+((a+2b)/a)+1+((−2a−b)/b)+1−3) ×((a/(a+2b))+1+(b/(−2a−b))+1+((−a−b)/(a−b))+1−3) =(((a−b)/(−a−b))+1+((a+2b)/a)+1+((−2a−b)/b)+1−3) ×((a/(a+2b))+1+(b/(−2a−b))+1+((−a−b)/(a−b))+1−3) =(((−2b)/(−a−b))+((2a+2b)/a)+((−2a)/b)−3) ×(((2a+2b)/(a+2b))+((−2a)/(−2a−b))+((−2b)/(a−b))−3) Continue
$${a}+{b}+{c}=\mathrm{0}\:,\:\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right)=? \\ $$$$\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{a}+{b}+{c}−\mathrm{2}{b}−{c}}{{c}}+\frac{{a}+{b}+{c}−{a}−\mathrm{2}{c}}{{a}}+\frac{{a}+{b}+{c}−\mathrm{2}{a}−{b}}{{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{{a}+{b}+{c}−{a}−\mathrm{2}{c}}+\frac{{b}}{{a}+{b}+{c}−\mathrm{2}{a}−{b}}+\frac{{c}}{{a}+{b}+{c}−\mathrm{2}{b}−{c}}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{−\mathrm{2}{b}−{c}}{{c}}+\frac{−{a}−\mathrm{2}{c}}{{a}}+\frac{−\mathrm{2}{a}−{b}}{{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{−{a}−\mathrm{2}{c}}+\frac{{b}}{−\mathrm{2}{a}−{b}}+\frac{{c}}{−\mathrm{2}{b}−{c}}\right) \\ $$$${a}+{b}+{c}=\mathrm{0}\Rightarrow\mathrm{c}=−{a}−{b} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{−\mathrm{2}{b}−\left(−{a}−{b}\right)}{−{a}−{b}}+\frac{−{a}−\mathrm{2}\left(−{a}−{b}\right)}{{a}}+\frac{−\mathrm{2}{a}−{b}}{{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{−{a}−\mathrm{2}\left(−{a}−{b}\right)}+\frac{{b}}{−\mathrm{2}{a}−{b}}+\frac{−{a}−{b}}{−\mathrm{2}{b}−\left(−{a}−{b}\right.}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{\left.−\mathrm{2}{b}+{a}+{b}\right)}{−{a}−{b}}+\frac{\left.−{a}+\mathrm{2}{a}+\mathrm{2}{b}\right)}{{a}}+\frac{−\mathrm{2}{a}−{b}}{{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{\left.−{a}+\mathrm{2}{a}+\mathrm{2}{b}\right)}+\frac{{b}}{−\mathrm{2}{a}−{b}}+\frac{−{a}−{b}}{−\mathrm{2}{b}+{a}+{b}}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{a}−{b}}{−{a}−{b}}+\mathrm{1}+\frac{{a}+\mathrm{2}{b}}{{a}}+\mathrm{1}+\frac{−\mathrm{2}{a}−{b}}{{b}}+\mathrm{1}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{{a}+\mathrm{2}{b}}+\mathrm{1}+\frac{{b}}{−\mathrm{2}{a}−{b}}+\mathrm{1}+\frac{−{a}−{b}}{{a}−{b}}+\mathrm{1}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{a}−{b}}{−{a}−{b}}+\mathrm{1}+\frac{{a}+\mathrm{2}{b}}{{a}}+\mathrm{1}+\frac{−\mathrm{2}{a}−{b}}{{b}}+\mathrm{1}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{{a}}{{a}+\mathrm{2}{b}}+\mathrm{1}+\frac{{b}}{−\mathrm{2}{a}−{b}}+\mathrm{1}+\frac{−{a}−{b}}{{a}−{b}}+\mathrm{1}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\frac{−\mathrm{2}{b}}{−{a}−{b}}+\frac{\mathrm{2}{a}+\mathrm{2}{b}}{{a}}+\frac{−\mathrm{2}{a}}{{b}}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\frac{\mathrm{2}{a}+\mathrm{2}{b}}{{a}+\mathrm{2}{b}}+\frac{−\mathrm{2}{a}}{−\mathrm{2}{a}−{b}}+\frac{−\mathrm{2}{b}}{{a}−{b}}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\mathrm{Continue} \\ $$

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