Question

If a, b, c are positive real numbers such that a+b-c/c=a-b+c/b=-a+b+c/a, find the value of (a+b) (b+c) (c+a) /abc.​

Answer 1

Answer:

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Answer 2

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$\sf{let}$

$\sf \dfrac{a + b + c}{c} = \dfrac{a - b + c}{b} = \dfrac{ - a + b + c}{a} = x$

$\sf \dfrac{a + b - c}{c} = x , \dfrac{a - b + c}{b} = x \: , \dfrac{ - a + b + c}{a} = x$

$\sf \dfrac{a + b - c}{c} = x , \dfrac{a - b + c}{b} = x \: , \dfrac{ - a + b + c}{a} = x$

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$\sf ⇰by \: adding \: we \: get$

$\begin{gathered}\sf a + \cancel b - \cancel c = cx \\ \sf \cancel a - b + c = bx \\ \sf - \cancel a + b + c = ax \\ \sf a + b + c = cx + bx + ax\end{gathered}$

$\begin{gathered}\sf ➱ a + b + c = x(a + b + c) \\ \sf \:\dfrac {\cancel a + \cancel b + \cancel c}{ \cancel a + \cancel b + \cancel c} = x\end{gathered}$

$\sf ⠀⠀ ⠀ ⠀⠀ ⠀⠀x = 1⠀⠀⠀⠀⠀⠀⠀x=1$

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$\begin{gathered}\sf ⇰by \: substituting \: the \: values \: of \\ \sf x \: in \: equation \: (1)\end{gathered} </p><p>$

$\begin{gathered}\sf ➯a + b - c = c(1) \\ \: a - b + c = b(1) \\ \: - a + b + c = a(1) \\ \sf a + b = x \: \\ a + c = 2b \: \\ b + c = 2a \\ \sf \: (a + b)(b + c)(c + a) = 2c \times 2b \times 2a\end{gathered}$

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$\sf ⇰divide \: both \: sides \: by \: abc$

$\sf ⇒ \dfrac{(a +b )( b+ c)(c +a )}{abc} = \dfrac{8\cancel a\cancel b\cancel c}{\cancel a \cancel b \cancel c}</p><p>$

$\sf ⇒ \dfrac{(a + b)(b + c)( c+a )}{abc} = 8$

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