Find the tangent to $y=x^3$ at $(-1,-1)$.

For the curve $y=x^3,$ its derivative is $\frac{dy}{dx}=3x^2.$

At $x=-1$, the slope is $m=3(-1)^2=3.$

Using point-slope form at $(-1,-1)$: $y+1=3(x+1).$ So the tangent is $y=3x+2.$

Find the points of intersection of the curve and the tangent.

Solve $x^3=3x+2.$ That is, $x^3-3x-2=0.$

Factorizing: $x^3-3x-2=(x+1)^2(x-2).$

Hence the intersection points are at $x=-1 \quad \text{and} \quad x=2.$

So the enclosed region lies between $x=-1$ and $x=2$.

Determine which graph is above the other in $[-1,2]$.

Consider $x^3-(3x+2)=x^3-3x-2=(x+1)^2(x-2).$

On the interval $[-1,2]$, we have $(x+1)^2\ge 0$ and $(x-2)\le 0$. Therefore, $x^3-(3x+2)\le 0.$

So the line $y=3x+2$ lies above the curve $y=x^3$ on this interval.

Set up the area integral.

Area enclosed is $A=\int_{-1}^{2}\big[(3x+2)-x^3\big]dx.$

Evaluate the integral.

$A=\int_{-1}^{2}(3x+2-x^3)dx$ $=\left[\frac{3x^2}{2}+2x-\frac{x^4}{4}\right]_{-1}^{2}.$

At $x=2$: $\frac{3(2)^2}{2}+2(2)-\frac{(2)^4}{4}=\frac{12}{2}+4-\frac{16}{4}=6+4-4=6.$

At $x=-1$: $\frac{3(-1)^2}{2}+2(-1)-\frac{(-1)^4}{4}=\frac{3}{2}-2-\frac{1}{4}$ $=\frac{6-8-1}{4}=-\frac{3}{4}.$

Therefore, $A=6-\left(-\frac{3}{4}\right)=\frac{24}{4}+\frac{3}{4}=\frac{27}{4}.$

Match with the options.

$\frac{27}{4}$ corresponds to Option C.