1. **State the problem:** Simplify the expression $$3\sqrt{8x^2y} + 5x\sqrt{2xy^2} - x\sqrt{32y}$$. 2. **Recall the rule:** The square root of a product can be separated as $$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$. 3. **Simplify each term:** - For $$3\sqrt{8x^2y}$$: $$8 = 4 \times 2$$, so $$\sqrt{8x^2y} = \sqrt{4 \times 2 \times x^2 \times y} = \sqrt{4} \times \sqrt{2} \times \sqrt{x^2} \times \sqrt{y} = 2 \times \sqrt{2} \times x \times \sqrt{y} = 2x\sqrt{2y}$$. Thus, $$3\sqrt{8x^2y} = 3 \times 2x \sqrt{2y} = 6x\sqrt{2y}$$. - For $$5x\sqrt{2xy^2}$$: $$\sqrt{2xy^2} = \sqrt{2} \times \sqrt{x} \times \sqrt{y^2} = \sqrt{2} \times \sqrt{x} \times y = y\sqrt{2x}$$. So, $$5x\sqrt{2xy^2} = 5x \times y \sqrt{2x} = 5xy\sqrt{2x}$$. - For $$x\sqrt{32y}$$: $$32 = 16 \times 2$$, so $$\sqrt{32y} = \sqrt{16 \times 2 \times y} = \sqrt{16} \times \sqrt{2} \times \sqrt{y} = 4 \times \sqrt{2} \times \sqrt{y} = 4\sqrt{2y}$$. Thus, $$x\sqrt{32y} = x \times 4\sqrt{2y} = 4x\sqrt{2y}$$. 4. **Rewrite the expression:** $$6x\sqrt{2y} + 5xy\sqrt{2x} - 4x\sqrt{2y}$$. 5. **Combine like terms:** The terms $$6x\sqrt{2y}$$ and $$-4x\sqrt{2y}$$ are like terms: $$6x\sqrt{2y} - 4x\sqrt{2y} = (6x - 4x)\sqrt{2y} = 2x\sqrt{2y}$$. 6. **Final simplified expression:** $$2x\sqrt{2y} + 5xy\sqrt{2x}$$. This is the fully simplified form because the two terms have different radicals and cannot be combined further.