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Question Number 60984 by naka3546 last updated on 28/May/19
(a/(a−b)) + (b/(b−c)) + (c/(c−a)) = 4 ab^2 + bc^2 + abc + ca^2 = a^2 b + b^2 c + c^2 a ((a/(a−b)))^3 + ((b/(b−c)))^3 + ((c/(c−a)))^3 = ?
$$\frac{{a}}{{a}−{b}}\:\:+\:\:\frac{{b}}{{b}−{c}}\:\:+\:\:\frac{{c}}{{c}−{a}}\:\:=\:\:\mathrm{4} \\ $$$${ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{abc}\:+\:{ca}^{\mathrm{2}} \:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\left(\frac{{a}}{{a}−{b}}\right)^{\mathrm{3}} \:\:+\:\:\left(\frac{{b}}{{b}−{c}}\right)^{\mathrm{3}} \:\:+\:\:\left(\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:\:=\:\:? \\ $$$$ \\ $$
Commented by naka3546 last updated on 28/May/19
73 ?
$$\mathrm{73}\:? \\ $$
Commented by Prithwish sen last updated on 28/May/19
ab^2 +bc^2 +abc+ca^2 =a^2 b+b^2 c+c^2 a ab(a−b)+bc(b−c)+ac(c−a)=abc (((a−b))/c)+(((b−c))/a)+(((c−a))/b)=1 ((−[(b−c)+(c−a)])/c)+ (((b−c))/a)+(((c−a))/b)=1 (b−c)[(1/a)−(1/c)] +(c−a)[(1/b)−(1/c)] =1 (((b−c)(c−a))/c)[(1/a)−(1/b)]=1 ((abc)/((a−b)(b−c)(c−a)))=−1 Now let A=(a/((a−b))), B=(b/((b−c))), C=(c/((c−a))) ∴ A+B+C=4, ABC=−1 we have to calculate A^3 +B^3 +C^3 please help.
$$\mathrm{ab}^{\mathrm{2}} +\mathrm{bc}^{\mathrm{2}} +\mathrm{abc}+\mathrm{ca}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} \mathrm{b}+\mathrm{b}^{\mathrm{2}} \mathrm{c}+\mathrm{c}^{\mathrm{2}} \mathrm{a} \\ $$$$\mathrm{ab}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{bc}\left(\mathrm{b}−\mathrm{c}\right)+\mathrm{ac}\left(\mathrm{c}−\mathrm{a}\right)=\mathrm{abc} \\ $$$$\frac{\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{c}}+\frac{\left(\mathrm{b}−\mathrm{c}\right)}{\mathrm{a}}+\frac{\left(\mathrm{c}−\mathrm{a}\right)}{\mathrm{b}}=\mathrm{1} \\ $$$$\frac{−\left[\left(\mathrm{b}−\mathrm{c}\right)+\left(\mathrm{c}−\mathrm{a}\right)\right]}{\mathrm{c}}+\:\frac{\left(\mathrm{b}−\mathrm{c}\right)}{\mathrm{a}}+\frac{\left(\mathrm{c}−\mathrm{a}\right)}{\mathrm{b}}=\mathrm{1} \\ $$$$\left(\mathrm{b}−\mathrm{c}\right)\left[\frac{\mathrm{1}}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{c}}\right]\:+\left(\mathrm{c}−\mathrm{a}\right)\left[\frac{\mathrm{1}}{\mathrm{b}}−\frac{\mathrm{1}}{\mathrm{c}}\right]\:=\mathrm{1} \\ $$$$\frac{\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)}{\mathrm{c}}\left[\frac{\mathrm{1}}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{b}}\right]=\mathrm{1} \\ $$$$\frac{\mathrm{abc}}{\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)}=−\mathrm{1} \\ $$$$\mathrm{Now}\:\mathrm{let}\: \\ $$$$\mathrm{A}=\frac{\mathrm{a}}{\left(\mathrm{a}−\mathrm{b}\right)},\:\mathrm{B}=\frac{\mathrm{b}}{\left(\mathrm{b}−\mathrm{c}\right)},\:\mathrm{C}=\frac{\mathrm{c}}{\left(\mathrm{c}−\mathrm{a}\right)} \\ $$$$\therefore\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{4},\:\mathrm{ABC}=−\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{calculate} \\ $$$$\mathrm{A}^{\mathrm{3}} +\mathrm{B}^{\mathrm{3}} +\mathrm{C}^{\mathrm{3}} \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$$$ \\ $$
Commented by naka3546 last updated on 28/May/19
(a/(a−b)) + (b/(b−c)) + (c/(c−a)) = 4 ⇒ 3 + ((b/(a−b)) + (c/(b−c)) + (a/(c−a))) = 4 ⇔ ((b/(a−b)) + (c/(b−c)) + (a/(c−a))) = 1 ⇒ ((abc − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 ⇔ abc = ab^2 + bc^2 + ca^2 ... (i) abc + ab^2 + bc^2 + ca^2 = a^2 b + b^2 c + c^2 a ⇔ a^2 b + b^2 c + c^2 a = 2abc ... (ii) ⇔ (a−b)(b−c)(c−a) = −abc ... (iii) ((a/(a−b)))^3 + ((b/(b−c)))^3 + ((c/(c−a)))^3 = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 − 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))(((ab)/((a−b)(b−c))) + ((bc)/((b−c)(c−a))) + ((ca)/((c−a)(a−b)))) + 3(((abc)/((a−b)(b−c)(c−a)))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 − 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))(− ((c−a)/c) − ((a−b)/a) − ((b−c)/b)) + 3(((abc)/((a−b)(b−c)(c−a)))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( ((c−a)/c) + ((a−b)/a) + ((b−c)/b)) + 3(((abc)/(−abc))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − ((a/c) + (b/a) + (c/b))) + 3(((abc)/(−abc))) = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − (((a^2 b + b^2 c + c^2 a)/(abc)))) − 3 = ((a/(a−b)) + (b/(b−c)) + (c/(c−a)))^3 + 3((a/(a−b)) + (b/(b−c)) + (c/(c−a)))( 3 − (((2abc)/(abc)))) − 3 = (4)^3 + 3(4)( 3 − 2) − 3 = 64 + 12 − 3 = 73
$$\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\:\:=\:\:\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{3}\:+\:\left(\frac{{b}}{{a}−{b}}\:+\:\frac{{c}}{{b}−{c}}\:+\:\frac{{a}}{{c}−{a}}\right)\:\:=\:\:\mathrm{4} \\ $$$$\Leftrightarrow\:\:\left(\frac{{b}}{{a}−{b}}\:+\:\frac{{c}}{{b}−{c}}\:+\:\frac{{a}}{{c}−{a}}\right)\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{abc}\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}{\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}\:=\:\mathrm{1} \\ $$$$\Leftrightarrow\:\:{abc}\:\:=\:\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:\:\:…\:\:\left({i}\right) \\ $$$$ \\ $$$${abc}\:\:+\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\Leftrightarrow\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\:\:=\:\:\mathrm{2}{abc}\:\:\:…\:\left({ii}\right) \\ $$$$ \\ $$$$\Leftrightarrow\:\:\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\:\:=\:\:−{abc}\:\:\:…\:\left({iii}\right) \\ $$$$ \\ $$$$\left(\frac{{a}}{{a}−{b}}\right)^{\mathrm{3}} \:+\:\left(\frac{{b}}{{b}−{c}}\right)^{\mathrm{3}} \:+\:\left(\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \: \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:−\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(\frac{{ab}}{\left({a}−{b}\right)\left({b}−{c}\right)}\:+\:\frac{{bc}}{\left({b}−{c}\right)\left({c}−{a}\right)}\:+\:\frac{{ca}}{\left({c}−{a}\right)\left({a}−{b}\right)}\right)\:+\:\mathrm{3}\left(\frac{{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\right) \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:−\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(−\:\frac{{c}−{a}}{{c}}\:−\:\frac{{a}−{b}}{{a}}\:−\:\frac{{b}−{c}}{{b}}\right)\:+\:\mathrm{3}\left(\frac{{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\right) \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:+\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(\:\frac{{c}−{a}}{{c}}\:+\:\frac{{a}−{b}}{{a}}\:+\:\frac{{b}−{c}}{{b}}\right)\:+\:\mathrm{3}\left(\frac{{abc}}{−{abc}}\right) \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:+\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(\:\mathrm{3}\:−\:\left(\frac{{a}}{{c}}\:+\:\frac{{b}}{{a}}\:+\:\frac{{c}}{{b}}\right)\right)\:+\:\mathrm{3}\left(\frac{{abc}}{−{abc}}\right) \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:+\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(\:\mathrm{3}\:−\:\left(\frac{{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}}{{abc}}\right)\right)\:−\:\mathrm{3} \\ $$$$=\:\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)^{\mathrm{3}} \:+\:\mathrm{3}\left(\frac{{a}}{{a}−{b}}\:+\:\frac{{b}}{{b}−{c}}\:+\:\frac{{c}}{{c}−{a}}\right)\left(\:\mathrm{3}\:−\:\left(\frac{\mathrm{2}{abc}}{{abc}}\right)\right)\:−\:\mathrm{3} \\ $$$$=\:\left(\mathrm{4}\right)^{\mathrm{3}} \:+\:\mathrm{3}\left(\mathrm{4}\right)\left(\:\mathrm{3}\:−\:\mathrm{2}\right)\:−\:\mathrm{3} \\ $$$$=\:\:\mathrm{64}\:+\:\mathrm{12}\:−\:\mathrm{3} \\ $$$$=\:\:\mathrm{73} \\ $$
Commented by MJS last updated on 28/May/19
how do you get (i)? (b/(a−b))+(c/(b−c))+(a/(c−a))≠((abc−(a^2 b+b^2 c+c^2 a))/((ab^2 +bc^2 +ca^2 )−(a^2 b+b^2 c+c^2 a)))
$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\left({i}\right)? \\ $$$$\frac{{b}}{{a}−{b}}+\frac{{c}}{{b}−{c}}+\frac{{a}}{{c}−{a}}\neq\frac{{abc}−\left({a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}\right)}{\left({ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}\right)} \\ $$
Commented by Prithwish sen last updated on 28/May/19
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Commented by naka3546 last updated on 28/May/19
ab^2 + bc^2 + ca^2 + abc = a^2 b + b^2 c + c^2 a ⇒ (a+b+c)(ab+bc+ca) − (2abc + ab^2 + bc^2 + ca^2 ) = a^2 b + b^2 c + c^2 a ⇔ (a+b+c)(ab+bc+ca) = 2(a^2 b + b^2 c + c^2 a) (b/(a−b)) + (c/(b−c)) + (a/(c−a)) = 1 ⇒ ((b(b−c)(c−a) + c(a−b)(c−a) + a(a−b)(b−c))/((a−b)(b−c)(c−a))) = 1 ⇒ ((b^2 c − bc^2 − ab^2 + abc + c^2 a − bc^2 − ca^2 + abc + a^2 b − ab^2 − ca^2 + abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((a^2 b + b^2 c + c^2 a) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((ab^2 + bc^2 + ca^2 + abc) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((4abc) − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 4abc = ab^2 + bc^2 + ca^2 ⇔ a^2 b + b^2 c + c^2 a = 5abc May be there are mistakes . Typo , sir.
$${ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:+\:{abc}\:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\Rightarrow\:\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)\:−\:\left(\mathrm{2}{abc}\:+\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\Leftrightarrow\:\:\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)\:\:=\:\:\mathrm{2}\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right) \\ $$$$ \\ $$$$\frac{{b}}{{a}−{b}}\:+\:\frac{{c}}{{b}−{c}}\:+\:\frac{{a}}{{c}−{a}}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{b}\left({b}−{c}\right)\left({c}−{a}\right)\:+\:{c}\left({a}−{b}\right)\left({c}−{a}\right)\:+\:{a}\left({a}−{b}\right)\left({b}−{c}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{b}^{\mathrm{2}} {c}\:−\:{bc}^{\mathrm{2}} \:−\:{ab}^{\mathrm{2}} \:+\:{abc}\:+\:{c}^{\mathrm{2}} {a}\:−\:{bc}^{\mathrm{2}} \:−\:{ca}^{\mathrm{2}} \:+\:{abc}\:+\:{a}^{\mathrm{2}} {b}\:−\:{ab}^{\mathrm{2}} \:−\:{ca}^{\mathrm{2}} \:+\:{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)\:−\:\mathrm{2}\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:+\:\mathrm{3}{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:+\:{abc}\right)\:−\:\mathrm{2}\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:+\:\mathrm{3}{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left(\mathrm{4}{abc}\right)\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}{\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{4}{abc}\:\:=\:\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\:\:=\:\:\mathrm{5}{abc} \\ $$$${May}\:\:{be}\:\:{there}\:\:{are}\:\:{mistakes}\:. \\ $$$${Typo}\:,\:\:{sir}. \\ $$
Commented by naka3546 last updated on 28/May/19
ab^2 + bc^2 + ca^2 + abc = a^2 b + b^2 c + c^2 a ⇒ (a+b+c)(ab+bc+ca) − (2abc + ab^2 + bc^2 + ca^2 ) = a^2 b + b^2 c + c^2 a ⇔ (a+b+c)(ab+bc+ca) = 2(a^2 b + b^2 c + c^2 a) (b/(a−b)) + (c/(b−c)) + (a/(c−a)) = 1 ⇒ ((b(b−c)(c−a) + c(a−b)(c−a) + a(a−b)(b−c))/((a−b)(b−c)(c−a))) = 1 ⇒ ((b^2 c − bc^2 − ab^2 + abc + c^2 a − bc^2 − ca^2 + abc + a^2 b − ab^2 − ca^2 + abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((a^2 b + b^2 c + c^2 a) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((ab^2 + bc^2 + ca^2 + abc) − 2(ab^2 + bc^2 + ca^2 ) + 3abc)/((a−b)(b−c)(c−a))) = 1 ⇒ (((4abc) − (a^2 b + b^2 c + c^2 a))/((ab^2 + bc^2 + ca^2 ) − (a^2 b + b^2 c + c^2 a))) = 1 4abc = ab^2 + bc^2 + ca^2 ⇔ a^2 b + b^2 c + c^2 a = 5abc May be there are mistakes . Typo , sir.
$${ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:+\:{abc}\:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\Rightarrow\:\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)\:−\:\left(\mathrm{2}{abc}\:+\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:\:=\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a} \\ $$$$\Leftrightarrow\:\:\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)\:\:=\:\:\mathrm{2}\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right) \\ $$$$ \\ $$$$\frac{{b}}{{a}−{b}}\:+\:\frac{{c}}{{b}−{c}}\:+\:\frac{{a}}{{c}−{a}}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{b}\left({b}−{c}\right)\left({c}−{a}\right)\:+\:{c}\left({a}−{b}\right)\left({c}−{a}\right)\:+\:{a}\left({a}−{b}\right)\left({b}−{c}\right)}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{b}^{\mathrm{2}} {c}\:−\:{bc}^{\mathrm{2}} \:−\:{ab}^{\mathrm{2}} \:+\:{abc}\:+\:{c}^{\mathrm{2}} {a}\:−\:{bc}^{\mathrm{2}} \:−\:{ca}^{\mathrm{2}} \:+\:{abc}\:+\:{a}^{\mathrm{2}} {b}\:−\:{ab}^{\mathrm{2}} \:−\:{ca}^{\mathrm{2}} \:+\:{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)\:−\:\mathrm{2}\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:+\:\mathrm{3}{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \:+\:{abc}\right)\:−\:\mathrm{2}\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:+\:\mathrm{3}{abc}}{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{\left(\mathrm{4}{abc}\right)\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}{\left({ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \right)\:−\:\left({a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\right)}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{4}{abc}\:\:=\:\:{ab}^{\mathrm{2}} \:+\:{bc}^{\mathrm{2}} \:+\:{ca}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\:{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{2}} {c}\:+\:{c}^{\mathrm{2}} {a}\:\:=\:\:\mathrm{5}{abc} \\ $$$${May}\:\:{be}\:\:{there}\:\:{are}\:\:{mistakes}\:. \\ $$$${Typo}\:,\:\:{sir}. \\ $$
Commented by naka3546 last updated on 28/May/19
37 ?
$$\mathrm{37}\:\:? \\ $$
Commented by MJS last updated on 28/May/19
I′ll have to look into it again...
$$\mathrm{I}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{look}\:\mathrm{into}\:\mathrm{it}\:\mathrm{again}… \\ $$
Answered by tanmay last updated on 28/May/19
(a/(a−b))+(b/(b−c))+(c/(c−a))=4 (1/(1−(b/a)))+(1/(1−(c/b)))+(1/(1−(a/c)))=4 u=(b/a) v=(c/b) w=(a/c) uvw=1 (1/(1−u))+(1/(1−v))+(1/(1−w))=4 (1/(1−u))−1+(1/(1−v))−1+(1/(1−w))−1=1 (u/(1−u))+(v/(1−v))+(w/(1−w))=1....(1) ((1/(1−u)))^3 +((1/(1−v)))^3 +((1/(1−w)))^3 =??? ab^2 +bc^2 +ca^2 +abc−a^2 b−b^2 c−c^2 a=0 ab(b−a)+bc(c−b)+ca(a−c)+abc=0 a^2 ×b((b/a)−1)+b^2 ×c((c/b)−1)+c^2 a((a/c)−1)+abc=0 a^3 ×(b/a)((b/a)−1)+b^3 ×(c/b)((c/b)−1)+c^3 ×(a/c)((a/c)−1)+abc=0 a^2 b(u−1)+b^2 c(v−1)+c^2 a(w−1)+abc=0 wait... u=(b/a) v=(c/b) w=(a/c) uvw=1 u:v:w=(b/a):(c/b):(a/c)=b^2 c:ac^2 :a^2 b b^2 c=uk ac^2 =vk a^2 b=wk a^3 b^3 c^3 =uvwk^3 abc=(uvw)^(1/3) k abc=k wk(u−1)+uk(v−1)+vk(w−1)+k=0 uwk−wk+uvk−uk+vwk−vk+k=0
$$\frac{{a}}{{a}−{b}}+\frac{{b}}{{b}−{c}}+\frac{{c}}{{c}−{a}}=\mathrm{4} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\frac{{b}}{{a}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{{c}}{{b}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{{a}}{{c}}}=\mathrm{4} \\ $$$${u}=\frac{{b}}{{a}}\:\:{v}=\frac{{c}}{{b}}\:\:\:{w}=\frac{{a}}{{c}}\:\:\:{uvw}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}−{v}}+\frac{\mathrm{1}}{\mathrm{1}−{w}}=\mathrm{4} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{u}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{v}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{w}}−\mathrm{1}=\mathrm{1} \\ $$$$\frac{{u}}{\mathrm{1}−{u}}+\frac{{v}}{\mathrm{1}−{v}}+\frac{{w}}{\mathrm{1}−{w}}=\mathrm{1}….\left(\mathrm{1}\right) \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{1}−{v}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{1}−{w}}\right)^{\mathrm{3}} =??? \\ $$$${ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} +{abc}−{a}^{\mathrm{2}} {b}−{b}^{\mathrm{2}} {c}−{c}^{\mathrm{2}} {a}=\mathrm{0} \\ $$$${ab}\left({b}−{a}\right)+{bc}\left({c}−{b}\right)+{ca}\left({a}−{c}\right)+{abc}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} ×{b}\left(\frac{{b}}{{a}}−\mathrm{1}\right)+{b}^{\mathrm{2}} ×{c}\left(\frac{{c}}{{b}}−\mathrm{1}\right)+{c}^{\mathrm{2}} {a}\left(\frac{{a}}{{c}}−\mathrm{1}\right)+{abc}=\mathrm{0} \\ $$$${a}^{\mathrm{3}} ×\frac{{b}}{{a}}\left(\frac{{b}}{{a}}−\mathrm{1}\right)+{b}^{\mathrm{3}} ×\frac{{c}}{{b}}\left(\frac{{c}}{{b}}−\mathrm{1}\right)+{c}^{\mathrm{3}} ×\frac{{a}}{{c}}\left(\frac{{a}}{{c}}−\mathrm{1}\right)+{abc}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {b}\left({u}−\mathrm{1}\right)+{b}^{\mathrm{2}} {c}\left({v}−\mathrm{1}\right)+{c}^{\mathrm{2}} {a}\left({w}−\mathrm{1}\right)+{abc}=\mathrm{0} \\ $$$${wait}… \\ $$$${u}=\frac{{b}}{{a}}\:\:\:{v}=\frac{{c}}{{b}}\:\:\:{w}=\frac{{a}}{{c}}\:\:\:{uvw}=\mathrm{1} \\ $$$${u}:{v}:{w}=\frac{{b}}{{a}}:\frac{{c}}{{b}}:\frac{{a}}{{c}}={b}^{\mathrm{2}} {c}:{ac}^{\mathrm{2}} :{a}^{\mathrm{2}} {b} \\ $$$${b}^{\mathrm{2}} {c}={uk}\:\:\:{ac}^{\mathrm{2}} ={vk}\:\:\:{a}^{\mathrm{2}} {b}={wk} \\ $$$${a}^{\mathrm{3}} {b}^{\mathrm{3}} {c}^{\mathrm{3}} ={uvwk}^{\mathrm{3}} \\ $$$${abc}=\left({uvw}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {k} \\ $$$${abc}={k} \\ $$$${wk}\left({u}−\mathrm{1}\right)+{uk}\left({v}−\mathrm{1}\right)+{vk}\left({w}−\mathrm{1}\right)+{k}=\mathrm{0} \\ $$$${uwk}−{wk}+{uvk}−{uk}+{vwk}−{vk}+{k}=\mathrm{0} \\ $$
Commented by Prithwish sen last updated on 28/May/19
great thinking sir.
$$\mathrm{great}\:\mathrm{thinking}\:\mathrm{sir}. \\ $$
Commented by tanmay last updated on 28/May/19
k(uw+uv+vw−u−v−w)=0 k≠0 uw+uv+vw−u−v−k=0 u+v+k=uv+vw+uw wait...
$${k}\left({uw}+{uv}+{vw}−{u}−{v}−{w}\right)=\mathrm{0} \\ $$$${k}\neq\mathrm{0} \\ $$$${uw}+{uv}+{vw}−{u}−{v}−{k}=\mathrm{0} \\ $$$${u}+{v}+{k}={uv}+{vw}+{uw} \\ $$$${wait}… \\ $$
Answered by MJS last updated on 29/May/19
let b=pa ∧ c=qa (1) ⇒ (3−2q)p^2 +(3q^2 −3q−2)p+q(3−2q)=0 (2) ⇒ (1−q)p^2 +(q^2 +q−1)p+q(1−q)=0 ⇒ q=.198062 p= { ((.307979)),((.643104)) :} q=1.55496 p= { ((.307979)),((5.04892)) :} q=3.24698 p= { ((.643104)),((5.04892)) :} in all cases we get the result 25 btw (1/(.198...))=5.04...; (1/(.307...))=3.24...; (1/(.643...))=1.55...
$$\mathrm{let}\:{b}={pa}\:\wedge\:{c}={qa} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:\left(\mathrm{3}−\mathrm{2}{q}\right){p}^{\mathrm{2}} +\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{3}{q}−\mathrm{2}\right){p}+{q}\left(\mathrm{3}−\mathrm{2}{q}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:\left(\mathrm{1}−{q}\right){p}^{\mathrm{2}} +\left({q}^{\mathrm{2}} +{q}−\mathrm{1}\right){p}+{q}\left(\mathrm{1}−{q}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${q}=.\mathrm{198062}\:{p}=\begin{cases}{.\mathrm{307979}}\\{.\mathrm{643104}}\end{cases} \\ $$$${q}=\mathrm{1}.\mathrm{55496}\:{p}=\begin{cases}{.\mathrm{307979}}\\{\mathrm{5}.\mathrm{04892}}\end{cases} \\ $$$${q}=\mathrm{3}.\mathrm{24698}\:{p}=\begin{cases}{.\mathrm{643104}}\\{\mathrm{5}.\mathrm{04892}}\end{cases} \\ $$$$\mathrm{in}\:\mathrm{all}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{25} \\ $$$$\mathrm{btw}\:\frac{\mathrm{1}}{.\mathrm{198}…}=\mathrm{5}.\mathrm{04}…;\:\frac{\mathrm{1}}{.\mathrm{307}…}=\mathrm{3}.\mathrm{24}…;\:\frac{\mathrm{1}}{.\mathrm{643}…}=\mathrm{1}.\mathrm{55}… \\ $$

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