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$If a, b, c are positive real numbers such that \frac{a + b - c}{c} = \frac{a - b + c}{b} = \frac{b + c - a}{a} = N, then the value of \frac{(a + b) (b + c) (c + a)}{abc} is$

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$N − 1 3$

$3 N + 1$

$N + 1 3$

$N + 1 2$

answer is C.

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Detailed Solution

$a, b, c ∈ I R − 1$

$a + b − c c = a − b + c b = b + c − a a = N$

$⇒ a + b − c = N c ⇒ a + b = c N + 1 → 1$

$a − b + c b = N ⇒ a − b + c = N b$

$⇒ a + c = b N + 1 → 2$

$b + c − a a = N ⇒ b + c − a = N a$

$⇒ b + c = a N + 1 → 3$

$∴ a + b b + c c + a a b c = c N + 1 b N + 1 a N + 1 a . b . c$

$= N + 1 3$

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