Question

If $a,b,c$ are positive real numbers such that $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},$ find the value of $\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}.$

Answer 1

Given $a,b,c$ are positive real numbers so $a+b+c\ne 0$

Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k\left(say\right)$

$⟹a+b-c=kc⟹a+b=(k+1)c\cdots \left(1\right)$

Similarly $a+c=(k+1)b\cdots \left(2\right),b+c=(k+1)a\cdots \left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)⟹2(a+b+c)=(k+1)(a+b+c)⟹k=1$

Multiplying $\left(1\right),\left(2\right),\left(3\right)$

$⟹(a+b)(b+c)(c+a)=(k+1{)}^{3}abc$

$⟹\frac{(a+b)(b+c)(c+a)}{abc}=(k+1{)}^3=(1+1{)}^3=2^3=8$