1. **State the problem:** Find the area of the shaded region bounded by the curves $$y = 8 - 3x^2$$ and $$y = x^3 - 4x^2 - 6x + 8$$. 2. **Find the points of intersection:** Set the two functions equal to find the limits of integration: $$8 - 3x^2 = x^3 - 4x^2 - 6x + 8$$ Simplify: $$8 - 3x^2 - 8 + 4x^2 + 6x - x^3 = 0$$ $$-x^3 + x^2 + 6x = 0$$ Multiply both sides by $$-1$$: $$x^3 - x^2 - 6x = 0$$ Factor out $$x$$: $$x(x^2 - x - 6) = 0$$ Factor quadratic: $$x(x - 3)(x + 2) = 0$$ So, $$x = 0, 3, -2$$ are intersection points. 3. **Determine which curve is on top between the intersection points:** Check at $$x=1$$: $$y_1 = 8 - 3(1)^2 = 8 - 3 = 5$$ $$y_2 = 1^3 - 4(1)^2 - 6(1) + 8 = 1 - 4 - 6 + 8 = -1$$ So, $$y = 8 - 3x^2$$ is above $$y = x^3 - 4x^2 - 6x + 8$$ between $$x = -2$$ and $$x = 3$$. 4. **Set up the integral for the area:** $$\text{Area} = \int_{-2}^{3} \left[(8 - 3x^2) - (x^3 - 4x^2 - 6x + 8)\right] dx$$ Simplify the integrand: $$8 - 3x^2 - x^3 + 4x^2 + 6x - 8 = -x^3 + ( -3x^2 + 4x^2 ) + 6x + (8 - 8) = -x^3 + x^2 + 6x$$ 5. **Integrate:** $$\int_{-2}^{3} (-x^3 + x^2 + 6x) dx = \left[-\frac{x^4}{4} + \frac{x^3}{3} + 3x^2\right]_{-2}^{3}$$ 6. **Evaluate at the bounds:** At $$x=3$$: $$-\frac{3^4}{4} + \frac{3^3}{3} + 3(3^2) = -\frac{81}{4} + 9 + 27 = -\frac{81}{4} + 36 = \frac{-81 + 144}{4} = \frac{63}{4}$$ At $$x=-2$$: $$-\frac{(-2)^4}{4} + \frac{(-2)^3}{3} + 3(-2)^2 = -\frac{16}{4} - \frac{8}{3} + 3(4) = -4 - \frac{8}{3} + 12 = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$$ 7. **Calculate the area:** $$\text{Area} = \frac{63}{4} - \frac{16}{3} = \frac{189}{12} - \frac{64}{12} = \frac{125}{12}$$ **Final answer:** $$\boxed{\frac{125}{12}}$$