If ` alpha and beta ` are the roots of equation ` x^(2) - x + 1 = 0` , then which equation will have roots ` alpha^(3) and beta^(3)` ?

Options

A. `x^(2) + 2 x + 1 = 0 `
B. ` x^(2) - 2 x - 1 = 0 `
C. ` x^(2) + 3 x - 1 = 0 `
D. `x^(2) - 3 x + 1 = 0`

Answer

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

### Step 1: Identify the roots \( \alpha \) and \( \beta \)
The roots of the equation \( x^2 - x + 1 = 0 \) can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -1 \), and \( c = 1 \).

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = 1 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = 1 \)

### Step 3: Find \( \alpha^3 + \beta^3 \)
We can use the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) \]
We already know \( \alpha + \beta = 1 \) and we need to find \( \alpha^2 + \beta^2 \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1^2 - 2 \cdot 1 = 1 - 2 = -1 \]
Thus,
\[ \alpha^3 + \beta^3 = 1 \cdot (-1 - 1) = -2 \]

### Step 4: Find \( \alpha^3 \beta^3 \)
Using the identity:
\[ \alpha^3 \beta^3 = (\alpha \beta)^3 = 1^3 = 1 \]

### Step 5: Form the new quadratic equation
The quadratic equation with roots \( \alpha^3 \) and \( \beta^3 \) can be written as:
\[ x^2 - (\alpha^3 + \beta^3)x + \alpha^3 \beta^3 = 0 \]
Substituting the values we found:
\[ x^2 - (-2)x + 1 = 0 \]
This simplifies to:
\[ x^2 + 2x + 1 = 0 \]

### Final Answer
The required quadratic equation is:
\[ x^2 + 2x + 1 = 0 \]