If `alpha and beta` are the roots of the equation `x^(2) - 4x + 1 = 0`, then the value of `alpha^(3) + beta^(3)` is

Options

A. 76
B. 52
C. -52
D. -76

Answer

To find the value of \( \alpha^3 + \beta^3 \) where \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 4x + 1 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is \( x^2 - 4x + 1 = 0 \). Here, we have:
- \( a = 1 \)
- \( b = -4 \)
- \( c = 1 \)

### Step 2: Calculate the roots using the quadratic formula
The roots of the equation can be calculated using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 4}}{2} \]
\[ x = \frac{4 \pm \sqrt{12}}{2} \]
\[ x = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ x = 2 \pm \sqrt{3} \]
Thus, the roots are:
\[ \alpha = 2 + \sqrt{3}, \quad \beta = 2 - \sqrt{3} \]

### Step 3: Use the identity for the sum of cubes
We can use the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]

### Step 4: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
From Vieta's formulas, we know:
\[ \alpha + \beta = 4 \quad (\text{sum of roots}) \]
\[ \alpha \beta = 1 \quad (\text{product of roots}) \]

### Step 5: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the known values:
\[ \alpha^2 + \beta^2 = 4^2 - 2 \cdot 1 = 16 - 2 = 14 \]

### Step 6: Substitute values into the sum of cubes formula
Now we can substitute back into the sum of cubes formula:
\[ \alpha^3 + \beta^3 = (4)(14 - 1) = 4 \cdot 13 = 52 \]

### Final Answer
Thus, the value of \( \alpha^3 + \beta^3 \) is \( \boxed{52} \).

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