1. **Statement of the problem:** Prove that: $$\left(\frac{3^a}{3^b}\right)^{a+b} \times \left(\frac{3^b}{3^c}\right)^{b+c} \times \left(\frac{3^c}{3^a}\right)^{c+a} = 1$$ 2. **Recall the exponent and division rules:** - $\frac{x^m}{x^n} = x^{m-n}$ - $(x^m)^n = x^{mn}$ - Multiplying powers with the same base adds exponents. 3. **Rewrite each fraction inside the parentheses:** $$\left(3^{a-b}\right)^{a+b} \times \left(3^{b-c}\right)^{b+c} \times \left(3^{c-a}\right)^{c+a}$$ 4. **Apply the power of a power rule:** $$3^{(a-b)(a+b)} \times 3^{(b-c)(b+c)} \times 3^{(c-a)(c+a)}$$ 5. **Multiply the bases by adding exponents:** $$3^{(a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a)}$$ 6. **Expand each product:** - $(a-b)(a+b) = a^2 - b^2$ - $(b-c)(b+c) = b^2 - c^2$ - $(c-a)(c+a) = c^2 - a^2$ 7. **Sum the exponents:** $$a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$$ 8. **Therefore:** $$3^0 = 1$$ **Final answer:** $$\boxed{1}$$