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

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To find the area enclosed by the curves $$ y = x^3 $$ and $$ y = 3x - x^2 $$, follow these steps:

### 1. Find the Points of Intersection
Set the two equations equal to each other to find the points where the curves intersect:

$$
x^3 = 3x - x^2
$$

Rearrange the equation:

$$
x^3 + x^2 - 3x = 0 \
x(x^2 + x - 3) = 0
$$

So, the roots are:
$$
x = 0
$$
$$
x = \frac{-1 \pm \sqrt{13}}{2}
$$

Therefore, the points of intersection are:
$$
x = \frac{-1 - \sqrt{13}}{2}, \quad x = 0, \quad x = \frac{-1 + \sqrt{13}}{2}
$$

### 2. Determine Which Function is on Top
To determine which function is above the other between the points of intersection, pick test points in each interval.

- **Between $$ \frac{-1 - \sqrt{13}}{2} $$ and $$ 0 $$:**
  
  Choose $$ x = -1 $$:
  $$
  y = (-1)^3 = -1 \
  y = 3(-1) - (-1)^2 = -4
  $$
  Here, $$ y = x^3 > y = 3x - x^2 $$.

- **Between $$ 0 $$ and $$ \frac{-1 + \sqrt{13}}{2} $$:**
  
  Choose $$ x = 1 $$:
  $$
  y = 1^3 = 1 \
  y = 3(1) - 1^2 = 2
  $$
  Here, $$ y = 3x - x^2 > y = x^3 $$.

### 3. Set Up the Integral
The total area is the sum of the areas between the curves in each interval:

$$
\text{Area} = \int_{\frac{-1 - \sqrt{13}}{2}}^{0} \left( x^3 - (3x - x^2) \right) dx + \int_{0}^{\frac{-1 + \sqrt{13}}{2}} \left( (3x - x^2) - x^3 \right) dx
$$

### 4. Calculate the Integral
After simplifying and evaluating the integrals, we find:

$$
\text{Area} = \frac{73}{12}
$$

### Final Answer
The area enclosed by the curves is $$ \boxed{\dfrac{73}{12}} $$.
The area enclosed by the curves $$ y = x^3 $$ and $$ y = 3x - x^2 $$ is $$ \frac{73}{12} $$.

Find the region’s area that is enclosed by the curves and .

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