Step 1: Set up the partial fraction equation.
We are given that \[ \frac{x^3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}. \] We want to find $C$, the coefficient of $\frac{1}{x+2}$.
Step 2: Understand the cover-up method.
To isolate $C$, multiply both sides by $(x+2)$ and then substitute $x = -2$. This makes all terms except $C$ vanish because they contain $(x+2)$ in their numerators or denominators.
Step 3: Apply the cover-up to find C.
After multiplying by $(x+2)$ and setting $x = -2$: \[ C = \left.\frac{x^3}{(2x-1)(x-3)}\right|_{x=-2}. \]
Step 4: Evaluate the numerator at x = -2.
The numerator is $x^3 = (-2)^3 = -8$.
Step 5: Evaluate the denominator at x = -2.
We get $2x-1 = 2(-2)-1 = -5$ and $x-3 = -2-3 = -5$. So the denominator is $(-5)(-5) = 25$.
Step 6: Compute C and state the answer.
\[ C = \frac{-8}{25}. \] This matches option (3).
\[ \boxed{-\dfrac{8}{25}} \]