If `2x + (2)/(x) =1,` then the value of `x ^(3) + (1)/(x ^(3)) ` is

Options

A. `(13)/(8)`
B. ` -(11)/(8)`
C. `(11)/(8)`
D. `- (13)/(8)`

Answer

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

### Step 1: Simplify the original equation
Start with the equation:
\[ 2x + \frac{2}{x} = 1 \]
To eliminate the fraction, multiply both sides by \( x \) (assuming \( x \neq 0 \)):
\[ 2x^2 + 2 = x \]

### Step 2: Rearrange the equation
Rearranging gives us a standard quadratic form:
\[ 2x^2 - x + 2 = 0 \]

### Step 3: Use the quadratic formula
To solve the quadratic equation \( 2x^2 - x + 2 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -1 \), and \( c = 2 \). Plugging in these values:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ x = \frac{1 \pm \sqrt{1 - 16}}{4} \]
\[ x = \frac{1 \pm \sqrt{-15}}{4} \]

### Step 4: Identify the roots
Since the discriminant \( 1 - 16 = -15 \) is negative, the roots are complex:
\[ x = \frac{1 \pm i\sqrt{15}}{4} \]

### Step 5: Find \( x + \frac{1}{x} \)
To find \( x^3 + \frac{1}{x^3} \), we first need \( x + \frac{1}{x} \). We can calculate \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{4}{1 \pm i\sqrt{15}} \]
To rationalize, multiply numerator and denominator by the conjugate:
\[ \frac{1}{x} = \frac{4(1 \mp i\sqrt{15})}{(1)^2 + (15)} = \frac{4(1 \mp i\sqrt{15})}{16} = \frac{1 \mp i\sqrt{15}}{4} \]
Thus,
\[ x + \frac{1}{x} = \frac{1 + i\sqrt{15}}{4} + \frac{1 - i\sqrt{15}}{4} = \frac{2}{4} = \frac{1}{2} \]

### Step 6: Calculate \( x^3 + \frac{1}{x^3} \)
Using the identity:
\[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \]
Substituting \( x + \frac{1}{x} = \frac{1}{2} \):
\[ x^3 + \frac{1}{x^3} = \left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) \]
\[ = \frac{1}{8} - \frac{3}{2} \]
\[ = \frac{1}{8} - \frac{12}{8} = -\frac{11}{8} \]

Thus, the value of \( x^3 + \frac{1}{x^3} \) is:
\[ \boxed{-\frac{11}{8}} \]