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


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

$\frac{a}{b + c} = \frac{b}{a + c} = \frac{c}{a + b}; a \cdot b \cdot c = 12$ $(b + c) \cdot (a + c) \cdot (a + b) = ?$
Commented by MJS_new last updated on 05/Jun/21

$I found 2 possibilities$ $(1) a = b = c = \sqrt[3]{12} \Rightarrow answer is 96$ $(2) a < b < c \Rightarrow a = - \sqrt[3]{96} \land b = \sqrt[3]{12} - \sqrt[6]{486} \land c = \sqrt[3]{12} + \sqrt[6]{486} \Rightarrow answer is - 12$
Commented by iloveisrael last updated on 05/Jun/21

$y e s . . . i g o t 2 p o s s i b i l i t i e s$
Commented by mathdanisur last updated on 05/Jun/21

$t h a n k s s i r$
	Answered by Olaf_Thorendsen last updated on 05/Jun/21

	$\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}$ $\Rightarrow \frac{b + c}{a} = \frac{c + a}{b} = \frac{a + b}{c} (1)$ $Let s = a + b + c$ $(1) : \frac{s - a}{a} = \frac{s - b}{b} = \frac{s - c}{c}$ $\Rightarrow \frac{s}{a} = \frac{s}{b} = \frac{s}{c}$ $\frac{1}{a} = \frac{1}{b} = \frac{1}{c}$ $\Rightarrow a = b = c$ $a b c = 12 = a^{3} = b^{3} = c^{3}$ $(b + c) (a + c) (a + b) = {(2 a)}^{3} = 8 a^{3} = 96$
	Commented by mathdanisur last updated on 05/Jun/21

	$t h a n k s s i r$
	Answered by iloveisrael last updated on 05/Jun/21
	$Given a×b×c=12 and (a/(b+c)) = (b/(a+c)) = (c/(a+b)) then (b+c)(a+c)(a+b)=? let (a/(b+c)) =k ⇒ { ((a=k(b+c))),((b=k(a+c))),((c=k(a+b))) :} ⇒a+b+c=k(2a+2b+2c) ⇒(a+b+c)(1−2k)=0 ⇒a+b+c = 0 or k=(1/2) for k=(1/2) →abc = ((1/2))^3 (b+c)(a+c)(a+b) ⇒(b+c)(a+c)(a+b)=12×8=96 for a=−(b+c) ⇒(b/(a+c)) = (c/(a+b)) =−1 ⇒a+c=−b ∧a+b=−c then (b+c)(a+c)(a+b)=(−a)(−b)(−c) =−abc = −12$
	$G i v e n a \times b \times c = 12 a n d$ $\frac{a}{b + c} = \frac{b}{a + c} = \frac{c}{a + b} t h e n$ $(b + c) (a + c) (a + b) = ?$ $l e t \frac{a}{b + c} = k \Rightarrow {\begin{cases} a = k (b + c) \\ b = k (a + c) \\ c = k (a + b) \end{cases}$ $\Rightarrow a + b + c = k (2 a + 2 b + 2 c)$ $\Rightarrow (a + b + c) (1 - 2 k) = 0$ $\Rightarrow a + b + c = 0 o r k = \frac{1}{2}$ $f o r k = \frac{1}{2}$ $\to a b c = {(\frac{1}{2})}^{3} (b + c) (a + c) (a + b)$ $\Rightarrow (b + c) (a + c) (a + b) = 12 \times 8 = 96$ $f o r a = - (b + c)$ $\Rightarrow \frac{b}{a + c} = \frac{c}{a + b} = - 1$ $\Rightarrow a + c = - b \land a + b = - c$ $t h e n (b + c) (a + c) (a + b) = (- a) (- b) (- c)$ $= - a b c = - 12$
	Commented by mathdanisur last updated on 05/Jun/21

	$t h a n k s s i r$
	Answered by 1549442205PVT last updated on 05/Jun/21

	$\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = \frac{a + b + c}{a + b + b + c + c + a} = \frac{1}{2}$ $\Rightarrow \frac{a}{b + c} . \frac{b}{c + a} . \frac{c}{a + b} = {(\frac{1}{2})}^{3} = \frac{1}{8}$ $\Rightarrow (a + b) (b + c) (c + a) = 8 abc = 8.12 = 96$
	Commented by mathdanisur last updated on 05/Jun/21

	$t h a n k s s i r$
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