If `x^(2)-2 sqrt(5)x+1=0` then which of the following is the value of `x^(3)+(1)/(x^(3))?`

Options

A. 34
B. `14 sqrt(5)`
C. 46
D. `34 sqrt(5)`

Answer

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

### Step 1: Solve the quadratic equation
The given equation is:
\[ x^2 - 2\sqrt{5}x + 1 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2\sqrt{5} \), and \( c = 1 \).

### Step 2: Calculate the discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (-2\sqrt{5})^2 = 4 \cdot 5 = 20 \]
\[ 4ac = 4 \cdot 1 \cdot 1 = 4 \]
Thus, the discriminant is:
\[ 20 - 4 = 16 \]

### Step 3: Find the roots
Now substituting back into the quadratic formula:
\[ x = \frac{2\sqrt{5} \pm \sqrt{16}}{2 \cdot 1} = \frac{2\sqrt{5} \pm 4}{2} \]
This simplifies to:
\[ x = \sqrt{5} + 2 \quad \text{or} \quad x = \sqrt{5} - 2 \]

### Step 4: Find \( x + \frac{1}{x} \)
Next, we need to find \( x + \frac{1}{x} \). We can calculate \( \frac{1}{x} \) for both roots.

For \( x = \sqrt{5} + 2 \):
\[ \frac{1}{x} = \frac{1}{\sqrt{5} + 2} \cdot \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{\sqrt{5} - 2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \]
Thus,
\[ x + \frac{1}{x} = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \]

### Step 5: Find \( x^3 + \frac{1}{x^3} \)
We can use the identity:
\[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \]
Substituting \( x + \frac{1}{x} = 2\sqrt{5} \):
\[ x^3 + \frac{1}{x^3} = (2\sqrt{5})^3 - 3(2\sqrt{5}) \]
Calculating \( (2\sqrt{5})^3 \):
\[ (2\sqrt{5})^3 = 8 \cdot 5\sqrt{5} = 40\sqrt{5} \]
Now substituting back:
\[ x^3 + \frac{1}{x^3} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5} \]

### Final Answer
Thus, the value of \( x^3 + \frac{1}{x^3} \) is:
\[ \boxed{34\sqrt{5}} \]