If the area enclosed by the curve `y^(2)=4x` and `16y^(2)=5(x-1)^(3)` can be expressed in the form `(L sqrt(M))/(N)` where L and N are relatively prime and M is a prime, find the value of `(L+M+N)`.

Answer

To solve the problem of finding the area enclosed by the curves \( y^2 = 4x \) and \( 16y^2 = 5(x-1)^3 \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ y^2 = 4x \quad \text{(1)} \]
This represents a rightward-opening parabola.

The second curve is given by:
\[ 16y^2 = 5(x-1)^3 \quad \text{(2)} \]
This curve is also a function of \( y \) and can be rewritten as:
\[ y^2 = \frac{5}{16}(x-1)^3 \quad \text{(3)} \]

### Step 2: Find the points of intersection
To find the points of intersection, we set the right-hand sides of equations (1) and (3) equal to each other:
\[ 4x = \frac{5}{16}(x-1)^3 \]
Multiplying both sides by 16 to eliminate the fraction gives:
\[ 64x = 5(x-1)^3 \]
Expanding the right-hand side:
\[ 64x = 5(x^3 - 3x^2 + 3x - 1) \]
This simplifies to:
\[ 64x = 5x^3 - 15x^2 + 15x - 5 \]
Rearranging gives:
\[ 5x^3 - 15x^2 - 49x - 5 = 0 \quad \text{(4)} \]

### Step 3: Solve for \( x \)
To find the roots of equation (4), we can use the Rational Root Theorem or synthetic division. Testing \( x = 5 \):
\[ 5(5^3) - 15(5^2) - 49(5) - 5 = 625 - 375 - 245 - 5 = 0 \]
Thus, \( x = 5 \) is a root. We can factor \( (x - 5) \) out of the polynomial.

Using synthetic division, we find:
\[ 5x^3 - 15x^2 - 49x - 5 = (x - 5)(5x^2 + 10x + 1) \]
Now, we can solve \( 5x^2 + 10x + 1 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{100 - 20}}{10} = \frac{-10 \pm \sqrt{80}}{10} = \frac{-10 \pm 4\sqrt{5}}{10} = \frac{-1 \pm \frac{2\sqrt{5}}{5}}{1} \]
This gives us two additional roots, but we are primarily interested in the intersection points, which are \( x = 1 \) and \( x = 5 \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 5 \) can be calculated as:
\[ A = 2 \int_{1}^{5} \left( \sqrt{\frac{5}{16}(x-1)^3} - \sqrt{4x} \right) dx \]

### Step 5: Calculate the integral
Calculating the integral:
1. The integral of \( \sqrt{4x} = 2\sqrt{x} \) from 1 to 5:
\[ \int 2\sqrt{x} \, dx = \frac{4}{3} x^{3/2} \bigg|_1^5 = \frac{4}{3} (5^{3/2} - 1^{3/2}) = \frac{4}{3} (5\sqrt{5} - 1) \]

2. The integral of \( \sqrt{\frac{5}{16}(x-1)^3} \):
\[ \int \sqrt{\frac{5}{16}} (x-1)^{3/2} \, dx = \frac{\sqrt{5}}{4} \cdot \frac{2}{5} (x-1)^{5/2} \bigg|_1^5 = \frac{\sqrt{5}}{10} \left( (5-1)^{5/2} - 0 \right) = \frac{\sqrt{5}}{10} \cdot 32 = \frac{32\sqrt{5}}{10} = \frac{16\sqrt{5}}{5} \]

Combining these results gives:
\[ A = 2 \left( \frac{4}{3}(5\sqrt{5} - 1) - \frac{16\sqrt{5}}{5} \right) \]
Simplifying this expression leads to:
\[ A = \frac{104\sqrt{5}}{15} \]

### Step 6: Express in the form \( \frac{L\sqrt{M}}{N} \)
Here, we have \( L = 104 \), \( M = 5 \), and \( N = 15 \). Since \( L \) and \( N \) are relatively prime, we can find:
\[ L + M + N = 104 + 5 + 15 = 124 \]

### Final Answer
Thus, the value of \( L + M + N \) is:
\[ \boxed{124} \]