Prove: \[ \left(\frac{3^a}{3^b}\right)^{a+b} \left(\frac{3^b}{3^c}\right)^{b+c} \left(\frac{3^c}{3^a}\right)^{c+a} = 1 \]
Proof
\[ = (3^{a-b})^{a+b}(3^{b-c})^{b+c}(3^{c-a})^{c+a} \]
\[ = 3^{(a-b)(a+b) + (b-c)(b+c) + (c-a)(c+a)} \]
\[ = 3^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)} \]
\[ = 3^0 \]
\[ = 1 \]