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Question Number 161356 by naka3546 last updated on 17/Dec/21
a + b + c = 0    Find  the  value  of     (((b−c)/a) + ((c−a)/b) + ((a−b)/c))((a/(b−c)) + (b/(c−a)) + (c/(a−b))) .
$${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0} \\ $$$$ \\ $$$${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\left(\frac{{b}−{c}}{{a}}\:+\:\frac{{c}−{a}}{{b}}\:+\:\frac{{a}−{b}}{{c}}\right)\left(\frac{{a}}{{b}−{c}}\:+\:\frac{{b}}{{c}−{a}}\:+\:\frac{{c}}{{a}−{b}}\right)\:. \\ $$
Commented by MJS_new last updated on 17/Dec/21
9
$$\mathrm{9} \\ $$
Commented by naka3546 last updated on 17/Dec/21
Show  your  workings, please,  sir.
$${Show}\:\:{your}\:\:{workings},\:{please},\:\:{sir}. \\ $$
Answered by MJS_new last updated on 17/Dec/21
c=−(a+b)  ⇒  ((b−c)/a)+((c−a)/b)++((a−b)/c)=−(((a−b)(a+2b)(2a+b))/(a(a+b)b))  (a/(b−c))+(b/(c−a))+(c/(a−b))=−((9a(a+b)b)/((a−b)(a+2b)(2a+b)))  ⇒ answer is 9
$${c}=−\left({a}+{b}\right) \\ $$$$\Rightarrow \\ $$$$\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}++\frac{{a}−{b}}{{c}}=−\frac{\left({a}−{b}\right)\left({a}+\mathrm{2}{b}\right)\left(\mathrm{2}{a}+{b}\right)}{{a}\left({a}+{b}\right){b}} \\ $$$$\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}=−\frac{\mathrm{9}{a}\left({a}+{b}\right){b}}{\left({a}−{b}\right)\left({a}+\mathrm{2}{b}\right)\left(\mathrm{2}{a}+{b}\right)} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{9} \\ $$
Commented by naka3546 last updated on 17/Dec/21
Thank  you,  sir.
$${Thank}\:\:{you},\:\:{sir}. \\ $$

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