The area of the region enclosed by the curve $$y = x^3$$ and its tangent at the point $$(-1, -1)$$ is

We need to find the area of the region enclosed by the curve $$y = x^3$$ and its tangent at the point $$(-1, -1)$$.

We begin by finding the equation of the tangent line at $$(-1, -1)$$. The derivative of $$y = x^3$$ is $$\frac{dy}{dx} = 3x^2$$, so at $$x = -1$$ the slope is $$m = 3(-1)^2 = 3$$. Using the point-slope form with point $$(-1, -1)$$, we have $$y - (-1) = 3(x - (-1))$$ which gives $$y + 1 = 3(x + 1)$$ and hence $$y = 3x + 2$$.

Next, we find the intersection points of $$y = x^3$$ and $$y = 3x + 2$$ by setting them equal: $$x^3 = 3x + 2$$, which leads to $$x^3 - 3x - 2 = 0$$. Since $$x = -1$$ is a point of tangency, $$(x + 1)^2$$ is a factor, and performing polynomial division yields $$x^3 - 3x - 2 = (x + 1)^2(x - 2) = 0$$. Therefore, the intersection points are $$x = -1$$ (double root, confirming tangency) and $$x = 2$$.

Then, to determine which function is on top in the interval $$[-1, 2]$$, we observe that at $$x = 0$$ the tangent line gives $$y = 2$$, while the curve gives $$y = 0$$. Thus the tangent line $$y = 3x + 2$$ lies above $$y = x^3$$ on $$[-1, 2]$$.

Finally, we compute the enclosed area by evaluating $$A = \int_{-1}^{2} \left[(3x + 2) - x^3\right] \, dx$$, which yields $$A = \left[\frac{3x^2}{2} + 2x - \frac{x^4}{4}\right]_{-1}^{2}$$. At $$x = 2$$,
$$\frac{3(4)}{2} + 2(2) - \frac{16}{4} = 6 + 4 - 4 = 6$$
and at $$x = -1$$,
$$\frac{3(1)}{2} + 2(-1) - \frac{1}{4} = \frac{3}{2} - 2 - \frac{1}{4} = \frac{6 - 8 - 1}{4} = -\frac{3}{4}$$. Therefore, $$A = 6 - \left(-\frac{3}{4}\right) = 6 + \frac{3}{4} = \frac{27}{4}$$.

The area of the enclosed region is $$\frac{27}{4}$$.
The correct answer is Option D.