If `((8x)/(2x^2+7x-2))=1, xgt0,` then what is the value of `x^3+1/x^3`

Options

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

Answer

To solve the equation \(\frac{8x}{2x^2 + 7x - 2} = 1\) for \(x > 0\) and find the value of \(x^3 + \frac{1}{x^3}\), we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation:

\[ 8x = 2x^2 + 7x - 2 \]

### Step 2: Simplifying the Equation
Move all terms to one side to form a quadratic equation:

\[ 2x^2 + 7x - 2 - 8x = 0 \]

This simplifies to:

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

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

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

Grouping the terms gives:

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

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

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

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

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

Thus, the valid solution is:

\[ x = 1 \]

### Step 5: Finding \(x^3 + \frac{1}{x^3}\)
Now, we need to find \(x^3 + \frac{1}{x^3}\). We can use the identity:

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

First, calculate \(x + \frac{1}{x}\):

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

Now substitute into the identity:

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

Calculating this gives:

\[ = 8 - 6 = 2 \]

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

\[ \boxed{2} \]