Math, asked by busarajesh1978, 11 months ago

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

Answers

Answered by malapatithulasidevi
5

Answer:

this may be the answer to the question hope it helps to you

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Answered by MysticalStar07
25

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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 a + b - c = cx \: , a - b + c = bx \: , - a + b + c = ax

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\sf ⇰by \: adding \: we \: get

\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

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

\sf  ⠀⠀ ⠀ ⠀⠀ ⠀⠀x = 1

 \orange {━━━━━━━━━━━━━━━}

\sf ⇰by \: substituting \: the \: values \: of  \\ \sf x \: in \: equation \: (1)

\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

 \red {━━━━━━━━━━━━━━━}

\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}

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

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