Find the region's area that is enclosed by the curves \( y = x^{3} \) and \( y = 3x - x^{2} \).

To find the area enclosed by the curves \( y = x^3 \) and \( y = 3x - x^2 \), we first need to determine the points of intersection of the two curves. We set the equations equal to each other: \[ x^3 = 3x - x^2 \] Rearranging, we find: \[ x^3 + x^2 - 3x = 0 \] Factoring out an \( x \): \[ x(x^2 + x - 3) = 0 \] This gives us one solution at \( x = 0 \). For the quadratic \( x^2 + x - 3 \), we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2} \] This gives us two more intersection points: \[ x = \frac{-1 + \sqrt{13}}{2}, \quad x = \frac{-1 - \sqrt{13}}{2} \] However, only \( x = \frac{-1 + \sqrt{13}}{2} \) is relevant since it's positive. Next, we identify the area between the curves from \( x = 0 \) to \( x = \frac{-1 + \sqrt{13}}{2} \). The area \( A \) can be calculated using the integral of the upper curve minus the lower curve: \[ A = \int_{0}^{\frac{-1 + \sqrt{13}}{2}} ((3x - x^2) - x^3) \, dx \] Calculating the integral: \[ A = \int_{0}^{\frac{-1 + \sqrt{13}}{2}} (3x - x^2 - x^3) \, dx \] We find: \[ = \left[ \frac{3}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 \right]_{0}^{\frac{-1 + \sqrt{13}}{2}} \] Evaluating at the bounds gives us the area. The final area will be a numerical value that can be calculated after plugging in the upper limit into our integrated formula and subtracting the lower limit's value (which is zero). This area represents the space between these two beautifully curved equations!