1. **State the problem:** Simplify the expression $$\frac{x^a y^{2a}}{x^{2a-1}} \div \frac{(x^2 y)^a}{x^{3+a} y^{-1}}$$. 2. **Rewrite the division as multiplication by the reciprocal:** $$\frac{x^a y^{2a}}{x^{2a-1}} \times \frac{x^{3+a} y^{-1}}{(x^2 y)^a}$$ 3. **Simplify each part:** - Simplify the first fraction: $$\frac{x^a y^{2a}}{x^{2a-1}} = x^{a - (2a - 1)} y^{2a} = x^{a - 2a + 1} y^{2a} = x^{-a + 1} y^{2a}$$ - Simplify the second fraction denominator: $$(x^2 y)^a = x^{2a} y^a$$ - So the second fraction is: $$\frac{x^{3+a} y^{-1}}{x^{2a} y^a} = x^{3+a - 2a} y^{-1 - a} = x^{3 - a} y^{-1 - a}$$ 4. **Multiply the two simplified expressions:** $$x^{-a + 1} y^{2a} \times x^{3 - a} y^{-1 - a} = x^{(-a + 1) + (3 - a)} y^{2a + (-1 - a)} = x^{-a + 1 + 3 - a} y^{2a - 1 - a} = x^{-2a + 4} y^{a - 1}$$ 5. **Final simplified expression:** $$\boxed{x^{-2a + 4} y^{a - 1}}$$ This means the expression simplifies to $x^{-2a + 4} y^{a - 1}$.