1. **State the problem:** We are given a vector function $$\mathbf{v} = \{6 Xx Yy + Zz^3, 3 Xx^3 - Zz, 3 Xx Zz^2 - Yy\}$$ and need to determine if it is irrotational or rotational by computing its curl. 2. **Recall the formula for curl:** For a vector field $$\mathbf{v} = \{P, Q, R\}$$, the curl is given by $$\nabla \times \mathbf{v} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$ 3. **Identify components:** $$P = 6 Xx Yy + Zz^3$$ $$Q = 3 Xx^3 - Zz$$ $$R = 3 Xx Zz^2 - Yy$$ 4. **Compute partial derivatives:** - $$\frac{\partial R}{\partial y} = \frac{\partial}{\partial Yy}(3 Xx Zz^2 - Yy) = -1$$ - $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial Zz}(3 Xx^3 - Zz) = -1$$ - $$\frac{\partial P}{\partial z} = \frac{\partial}{\partial Zz}(6 Xx Yy + Zz^3) = 3 Zz^2$$ - $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial Xx}(3 Xx Zz^2 - Yy) = 3 Zz^2$$ - $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial Xx}(3 Xx^3 - Zz) = 9 Xx^2$$ - $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial Yy}(6 Xx Yy + Zz^3) = 6 Xx$$ 5. **Calculate each component of the curl:** - $$\left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) = (-1) - (-1) = 0$$ - $$\left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) = 3 Zz^2 - 3 Zz^2 = 0$$ - $$\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) = 9 Xx^2 - 6 Xx = 3 Xx (3 Xx - 2)$$ 6. **Interpret the result:** The curl is $$\{0, 0, 3 Xx (3 Xx - 2)\}$$. 7. **Conclusion:** Since the curl is not the zero vector everywhere (it depends on $$Xx$$), the vector function is rotational except at points where $$3 Xx (3 Xx - 2) = 0$$ (i.e., $$Xx=0$$ or $$Xx=\frac{2}{3}$$). **Final answer:** The given vector function is rotational.