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 a parabola given by:
\[ y^2 = 4x \]
The second curve can be rewritten as:
\[ y^2 = \frac{5}{16}(x-1)^3 \]

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

### Step 3: Find roots of the cubic equation
Using trial and error or synthetic division, we find that \( x = 5 \) is a root. We can factor the cubic polynomial:
\[ 5(x - 5)(x^2 + ax + b) = 0 \]
By performing synthetic division, we find:
\[ 5(x - 5)(x^2 + 1) = 0 \]
The roots are \( x = 5 \) and \( x^2 + 1 = 0 \) (which gives complex roots).

### Step 4: Determine the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 5 \) can be computed as:
\[ A = 2 \int_0^5 \sqrt{4x} \, dx - 2 \int_1^5 \sqrt{\frac{5}{16}(x-1)^3} \, dx \]

### Step 5: Calculate the first integral
Calculating the first integral:
\[ \int_0^5 \sqrt{4x} \, dx = \int_0^5 2\sqrt{x} \, dx = 2 \cdot \frac{2}{3} x^{3/2} \bigg|_0^5 = \frac{4}{3} \cdot (5^{3/2}) = \frac{4 \cdot 5\sqrt{5}}{3} = \frac{20\sqrt{5}}{3} \]

### Step 6: Calculate the second integral
Calculating the second integral:
\[ \int_1^5 \sqrt{\frac{5}{16}(x-1)^3} \, dx = \sqrt{\frac{5}{16}} \int_1^5 (x-1)^{3/2} \, dx = \frac{\sqrt{5}}{4} \cdot \frac{2}{5} (x-1)^{5/2} \bigg|_1^5 \]
Evaluating this gives:
\[ = \frac{\sqrt{5}}{4} \cdot \frac{2}{5} \cdot (4^{5/2}) = \frac{\sqrt{5}}{4} \cdot \frac{2}{5} \cdot 32 = \frac{16\sqrt{5}}{5} \]

### Step 7: Combine the areas
Thus, the total area is:
\[ A = 2 \left( \frac{20\sqrt{5}}{3} \right) - 2 \left( \frac{16\sqrt{5}}{5} \right) \]
Calculating gives:
\[ = \frac{40\sqrt{5}}{3} - \frac{32\sqrt{5}}{5} \]
Finding a common denominator (15):
\[ = \frac{200\sqrt{5}}{15} - \frac{96\sqrt{5}}{15} = \frac{104\sqrt{5}}{15} \]

### Step 8: Express in the required form
The area can be expressed as:
\[ \frac{104\sqrt{5}}{15} \]
Here, \( L = 104 \), \( M = 5 \), and \( N = 15 \).

### Step 9: Find \( L + M + N \)
Finally, we calculate:
\[ L + M + N = 104 + 5 + 15 = 124 \]

Thus, the final answer is:
\[ \boxed{124} \]