1. **State the problem:** Multiply $$\sqrt{77a^3} \cdot \sqrt{35b^5}$$ and simplify assuming $$a \geq 0$$ and $$b \geq 0$$. 2. **Use the property of square roots:** $$\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$$. 3. **Apply the property:** $$\sqrt{77a^3} \cdot \sqrt{35b^5} = \sqrt{77a^3 \cdot 35b^5}$$ 4. **Multiply the constants and variables inside the root:** $$77 \cdot 35 = 2695$$ $$a^3 \cdot b^5 = a^3 b^5$$ So, $$\sqrt{2695 a^3 b^5}$$ 5. **Factor inside the root to simplify:** $$2695 = 5 \times 7^2 \times 11$$ $$a^3 = a^2 \cdot a$$ $$b^5 = b^4 \cdot b$$ Rewrite: $$\sqrt{5 \times 7^2 \times 11 \times a^2 \times a \times b^4 \times b}$$ 6. **Separate perfect squares from the rest:** $$\sqrt{7^2} = 7$$ $$\sqrt{a^2} = a$$ $$\sqrt{b^4} = b^2$$ So, $$\sqrt{2695 a^3 b^5} = 7 a b^2 \sqrt{5 \times 11 \times a \times b} = 7 a b^2 \sqrt{55 a b}$$ 7. **Final simplified form:** $$7 a b^2 \sqrt{55 a b}$$ This is the simplest form assuming $$a \geq 0$$ and $$b \geq 0$$.