If `(6x)/((2x^(2) + 5x - 2)) = 1, x gt 0`, then the value of `x^(3) + (1)/(x^(3))` is :

Options

A. A)`(3)/(8)sqrt(17)`
B. B)`(5sqrt(17))/(8)`
C. C)`(5sqrt(17))/(16)`
D. D)`(3)/(4)sqrt(17)`

Answer

To solve the equation \( \frac{6x}{2x^2 + 5x - 2} = 1 \) for \( x > 0 \) and find the value of \( x^3 + \frac{1}{x^3} \), we will follow these steps:

### Step 1: Simplify the Equation
Start by cross-multiplying to eliminate the fraction:

\[ 6x = 2x^2 + 5x - 2 \]

### Step 2: Rearrange the Equation
Rearranging the equation gives:

\[ 2x^2 + 5x - 2 - 6x = 0 \]

This simplifies to:

\[ 2x^2 - x - 2 = 0 \]

### Step 3: Factor the Quadratic Equation
Next, we can factor the quadratic equation \( 2x^2 - x - 2 = 0 \). We look for two numbers that multiply to \( 2 \times -2 = -4 \) and add to \( -1 \). The numbers \( -2 \) and \( 1 \) work:

\[ 2x^2 - 2x + x - 2 = 0 \]

Grouping gives:

\[ 2x(x - 1) + 1(x - 1) = 0 \]

Factoring out \( (x - 1) \):

\[ (2x + 1)(x - 1) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor to zero gives:

1. \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \) (not valid since \( x > 0 \))
2. \( x - 1 = 0 \) → \( x = 1 \)

### Step 5: Find \( x + \frac{1}{x} \)
Now that we have \( x = 1 \), we calculate:

\[ x + \frac{1}{x} = 1 + \frac{1}{1} = 2 \]

### Step 6: Find \( x^3 + \frac{1}{x^3} \)
Using the identity:

\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \]

Substituting \( x + \frac{1}{x} = 2 \):

\[ x^3 + \frac{1}{x^3} = 2^3 - 3 \cdot 2 = 8 - 6 = 2 \]

### Final Answer
Thus, the value of \( x^3 + \frac{1}{x^3} \) is:

\[ \boxed{2} \]