An algebra problem by Anuj Shikarkhane

$\dfrac {(a+b)(b+c)(c+a)}{abc} = \dfrac {(-c)(-a)(-b)}{abc} = \dfrac {-abc}{abc} = \boxed{-1}$

In these kind of questions, all the $a$ , $b$ and $c$ could be factorized.

Therefore, we can put any $a$ , $b$ and $c$ that satisfy $a+b+c=0$ and $abc\neq0$ ,

Let us say that $a=1$ , $b=1$ and $c=-2$ ,

$\large \frac{(a+b)(b+c)(c+a)}{abc}\\=\large \frac{[(1)+(1)][(1)+(-2)][(-2)+(1)]}{(1)(1)(-2)}\\=\large \frac{(2)(-1)(-1)}{-2}\\=\large \frac{2}{-2}\\=\boxed{-1}$