If $x=by+cz$, $y=cz+ax$, and $z=ax+by$, then the value of the expression:
$$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}$$

is equal to:

Given the set of equations:
1. $x=by+cz$
2. $y=cz+ax$
3. $z=ax+by$

Let's add $ax$ to both sides of equation (1):

$$x+ax=ax+by+cz$$

$$x(1+a)=ax+by+cz⟹\frac{1}{1+a}=\frac{x}{ax+by+cz}$$

Multiplying both sides by $a$:

$$\frac{a}{1+a}=\frac{ax}{ax+by+cz}$$

Similarly, by adding $by$ to both sides of equation (2) and $cz$ to both sides of equation (3), we get:

$$\frac{b}{1+b}=\frac{by}{ax+by+cz}$$

$$\frac{c}{1+c}=\frac{cz}{ax+by+cz}$$

Now, summing the three expressions together:

$$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=\frac{ax+by+cz}{ax+by+cz}=1$$