Let us subtract $1$ from the left and right parts of each equality and obtain:

$$b - 1 = \frac{a+c}{a+1} - 1 = \frac{a+c - a - 1}{a+1} = \frac{c-1}{a+1},$$

and analogously from the other two equalities. Because of the conditions of existence of expressions, none of the variables equals $-1$, so we can multiply these equalities by the corresponding denominator:
$$(b-1)(a+1) = c-1,$$

and similarly for the other variables. Then we multiply all the equalities:
$$(a-1)(a+1)(b-1)(b+1)(c-1)(c+1) = (a-1)(b-1)(c-1).$$

If, e.g., $c=1$, we immediately get that $b=1$, hence $a=1$. In this case $(a+1)(b+1)(c+1)=8$. If none of the variables is equal to $1$, then we can divide by $(a-1)(b-1)(c-1)$, and therefore obtain $(a+1)(b+1)(c+1)=1$. Note that the value is $1$ when $a=b=c=0$, and it is $8$ for $a=b=c=1$.