If $a+b+c=0$, then $\frac{1}{x^b+x^{-c}+1}+\frac{1}{x^c+x^{-a}+1}+\frac{1}{x^a+x^{-b}+1}$ is equal to:

Let the given expression be denoted as $S=T_1+T_2+T_3$, where:
$$T_1=\frac{1}{x^b+x^{-c}+1}$$

$$T_2=\frac{1}{x^c+x^{-a}+1}$$

$$T_3=\frac{1}{x^a+x^{-b}+1}$$

We are given that $a+b+c=0⟹c=-a-b$, $a=-b-c$, and $b=-a-c$.

Let's modify the terms to have a common denominator.

*For the first term ($T_1$):** Multiply numerator and denominator by $x^{-b}$:

$$T_1=\frac{x^{-b}}{x^{-b}(x^b+x^{-c}+1)}=\frac{x^{-b}}{1+x^{-b-c}+x^{-b}}$$

Since $-b-c=a$, this becomes:

$$T_1=\frac{x^{-b}}{1+x^a+x^{-b}}\text{— (1)}$$

*For the second term ($T_2$):** Multiply numerator and denominator by $x^a$:

$$T_2=\frac{x^a}{x^a(x^c+x^{-a}+1)}=\frac{x^a}{x^{a+c}+1+x^a}$$

Since $a+c=-b$, this becomes:

$$T_2=\frac{x^a}{x^{-b}+1+x^a}\text{— (2)}$$

*For the third term ($T_3$):** Keep it as it is:

$$T_3=\frac{1}{1+x^a+x^{-b}}\text{— (3)}$$

Now, add the three expressions together as they all share the exact same denominator ($1+x^a+x^{-b}$):

$$S=\frac{x^{-b}+x^a+1}{1+x^a+x^{-b}}=1$$