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Question Number 142740 by mathdanisur last updated on 04/Jun/21

(a/(b+c)) = (b/(a+c)) = (c/(a+b)) ; a∙b∙c = 12 (b+c)∙(a+c)∙(a+b) = ?

$$\frac{{a}}{{b}+{c}}\:=\:\frac{{b}}{{a}+{c}}\:=\:\frac{{c}}{{a}+{b}}\:\:;\:\:{a}\centerdot{b}\centerdot{c}\:=\:\mathrm{12} \\ $$$$\left({b}+{c}\right)\centerdot\left({a}+{c}\right)\centerdot\left({a}+{b}\right)\:=\:? \\ $$

Commented by MJS_new last updated on 05/Jun/21

I found 2 possibilities (1) a=b=c=((12))^(1/3) ⇒ answer is 96 (2) a<b<c ⇒ a=−((96))^(1/3) ∧b=((12))^(1/3) −((486))^(1/6) ∧c=((12))^(1/3) +((486))^(1/6) ⇒ answer is −12

$$\mathrm{I}\:\mathrm{found}\:\mathrm{2}\:\mathrm{possibilities} \\ $$$$\left(\mathrm{1}\right)\:{a}={b}={c}=\sqrt[{\mathrm{3}}]{\mathrm{12}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{96} \\ $$$$\left(\mathrm{2}\right)\:{a}<{b}<{c}\:\Rightarrow\:{a}=−\sqrt[{\mathrm{3}}]{\mathrm{96}}\wedge{b}=\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{6}}]{\mathrm{486}}\wedge{c}=\sqrt[{\mathrm{3}}]{\mathrm{12}}+\sqrt[{\mathrm{6}}]{\mathrm{486}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\mathrm{12} \\ $$

Commented by iloveisrael last updated on 05/Jun/21

yes... i got 2 possibilities

$${yes}...\:{i}\:{got}\:\mathrm{2}\:{possibilities}\: \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by Olaf_Thorendsen last updated on 05/Jun/21

(a/(b+c)) = (b/(c+a)) = (c/(a+b)) ⇒ ((b+c)/a) = ((c+a)/b) = ((a+b)/c) (1) Let s = a+b+c (1) : ((s−a)/a) = ((s−b)/b) = ((s−c)/c) ⇒ (s/a) = (s/b) = (s/c) (1/a) = (1/b) = (1/c) ⇒ a = b = c abc = 12 = a^3 = b^3 = c^3 (b+c)(a+c)(a+b) = (2a)^3 = 8a^3 = 96

$$\frac{{a}}{{b}+{c}}\:=\:\frac{{b}}{{c}+{a}}\:=\:\frac{{c}}{{a}+{b}} \\ $$$$\Rightarrow\:\frac{{b}+{c}}{{a}}\:=\:\frac{{c}+{a}}{{b}}\:=\:\frac{{a}+{b}}{{c}}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{s}\:=\:{a}+{b}+{c} \\ $$$$\left(\mathrm{1}\right)\::\:\frac{{s}−{a}}{{a}}\:=\:\frac{{s}−{b}}{{b}}\:=\:\frac{{s}−{c}}{{c}} \\ $$$$\Rightarrow\:\frac{{s}}{{a}}\:=\:\frac{{s}}{{b}}\:=\:\frac{{s}}{{c}} \\ $$$$\:\frac{\mathrm{1}}{{a}}\:=\:\frac{\mathrm{1}}{{b}}\:=\:\frac{\mathrm{1}}{{c}} \\ $$$$\Rightarrow\:{a}\:=\:{b}\:=\:{c} \\ $$$${abc}\:=\:\mathrm{12}\:=\:{a}^{\mathrm{3}} \:=\:{b}^{\mathrm{3}} \:=\:{c}^{\mathrm{3}} \\ $$$$\left({b}+{c}\right)\left({a}+{c}\right)\left({a}+{b}\right)\:=\:\left(\mathrm{2}{a}\right)^{\mathrm{3}} \:=\:\mathrm{8}{a}^{\mathrm{3}} \:=\:\mathrm{96} \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by iloveisrael last updated on 05/Jun/21

Given a×b×c=12 and (a/(b+c)) = (b/(a+c)) = (c/(a+b)) then (b+c)(a+c)(a+b)=? let (a/(b+c)) =k ⇒ { ((a=k(b+c))),((b=k(a+c))),((c=k(a+b))) :} ⇒a+b+c=k(2a+2b+2c) ⇒(a+b+c)(1−2k)=0 ⇒a+b+c = 0 or k=(1/2) for k=(1/2) →abc = ((1/2))^3 (b+c)(a+c)(a+b) ⇒(b+c)(a+c)(a+b)=12×8=96 for a=−(b+c) ⇒(b/(a+c)) = (c/(a+b)) =−1 ⇒a+c=−b ∧a+b=−c then (b+c)(a+c)(a+b)=(−a)(−b)(−c) =−abc = −12

$$\:{Given}\:{a}×{b}×{c}=\mathrm{12}\:{and}\: \\ $$$$\:\frac{{a}}{{b}+{c}}\:=\:\frac{{b}}{{a}+{c}}\:=\:\frac{{c}}{{a}+{b}}\:{then}\: \\ $$$$\:\left({b}+{c}\right)\left({a}+{c}\right)\left({a}+{b}\right)=? \\ $$$${let}\:\frac{{a}}{{b}+{c}}\:={k}\:\Rightarrow\begin{cases}{{a}={k}\left({b}+{c}\right)}\\{{b}={k}\left({a}+{c}\right)}\\{{c}={k}\left({a}+{b}\right)}\end{cases} \\ $$$$\Rightarrow{a}+{b}+{c}={k}\left(\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{2}{c}\right) \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left(\mathrm{1}−\mathrm{2}{k}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}+{c}\:=\:\mathrm{0}\:{or}\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${for}\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\rightarrow{abc}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \left({b}+{c}\right)\left({a}+{c}\right)\left({a}+{b}\right) \\ $$$$\Rightarrow\left({b}+{c}\right)\left({a}+{c}\right)\left({a}+{b}\right)=\mathrm{12}×\mathrm{8}=\mathrm{96} \\ $$$${for}\:{a}=−\left({b}+{c}\right) \\ $$$$\Rightarrow\frac{{b}}{{a}+{c}}\:=\:\frac{{c}}{{a}+{b}}\:=−\mathrm{1}\: \\ $$$$\Rightarrow{a}+{c}=−{b}\:\wedge{a}+{b}=−{c} \\ $$$${then}\:\left({b}+{c}\right)\left({a}+{c}\right)\left({a}+{b}\right)=\left(−{a}\right)\left(−{b}\right)\left(−{c}\right) \\ $$$$\:\:\:\:\:\:=−{abc}\:=\:−\mathrm{12} \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by 1549442205PVT last updated on 05/Jun/21

(a/(b+c))=(b/(c+a))=(c/(a+b))=((a+b+c)/(a+b+b+c+c+a))=(1/2) ⇒(a/(b+c)).(b/(c+a)).(c/(a+b))=((1/2))^3 =(1/8) ⇒(a+b)(b+c)(c+a)=8abc=8.12=96

$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{b}+\mathrm{c}+\mathrm{c}+\mathrm{a}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{b}}{\mathrm{c}+\mathrm{a}}.\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{b}+\mathrm{c}\right)\left(\mathrm{c}+\mathrm{a}\right)=\mathrm{8abc}=\mathrm{8}.\mathrm{12}=\mathrm{96} \\ $$

Commented by mathdanisur last updated on 05/Jun/21

thanks sir

$${thanks}\:{sir} \\ $$

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