The value of $\left(\dfrac{b}{x^a}\right)^{b+c} \cdot \left(\dfrac{c}{x^b}\right)^{c+a} \cdot \left(\dfrac{a}{x^c}\right)^{a+b}$-style identity in Example 13 simplifies to:

• a$0$
• b$1$
• c$abc$
• d$a + b + c$