1. **State the problem:** Solve the equation $3^{x+y} = \sqrt[3]{27}$ for $x+y$. 2. **Recall the properties of exponents and roots:** The cube root of a number $a$ is $a^{1/3}$, so $\sqrt[3]{27} = 27^{1/3}$. 3. **Simplify the right side:** Since $27 = 3^3$, we have $$\sqrt[3]{27} = (3^3)^{1/3} = 3^{3 \times \frac{1}{3}} = 3^1 = 3.$$ 4. **Rewrite the equation:** $$3^{x+y} = 3.$$ 5. **Since the bases are equal and nonzero, set the exponents equal:** $$x + y = 1.$$ **Final answer:** $$x + y = 1.$$