If `x^(2) - 3x + 1 = 0` , then the value of `x^(3) + (1)/(x^(3))` is equal to

Options

A. 87
B. 123
C. 135
D. 201

Answer

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

### Step 1: Solve the quadratic equation
We start with the equation:
\[ x^2 - 3x + 1 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -3\), and \(c = 1\).

### Step 2: Calculate the discriminant
Calculate the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]

### Step 3: Find the roots
Now substituting back into the quadratic formula:
\[ x = \frac{3 \pm \sqrt{5}}{2} \]

### Step 4: Find \(x + \frac{1}{x}\)
To find \(x + \frac{1}{x}\), we can use the identity:
\[ x + \frac{1}{x} = \frac{x^2 + 1}{x} \]
From our original equation \(x^2 - 3x + 1 = 0\), we can rearrange it to find:
\[ x^2 = 3x - 1 \]
Substituting this into the identity gives:
\[ x + \frac{1}{x} = \frac{(3x - 1) + 1}{x} = \frac{3x}{x} = 3 \]

### Step 5: Cube \(x + \frac{1}{x}\)
Now we can 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 \(x + \frac{1}{x} = 3\):
\[ x^3 + \frac{1}{x^3} = 3^3 - 3 \cdot 3 \]

### Step 6: Calculate the final value
Calculating this gives:
\[ x^3 + \frac{1}{x^3} = 27 - 9 = 18 \]

Thus, the value of \(x^3 + \frac{1}{x^3}\) is:
\[ \boxed{18} \]