If `alpha,beta` be the roots of the equation `x^(2)-x-1=0`, determine the value of i) `alpha^(2)+beta^(2) and `
(ii) `alpha^(3)+beta^(3)`.

Answer

To solve the problem, we need to find the values of \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - x - 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation**:
The given equation is \( x^2 - x - 1 = 0 \).
- Coefficient of \( x^2 \) (a) = 1
- Coefficient of \( x \) (b) = -1
- Constant term (c) = -1

2. **Find the sum and product of the roots**:
Using Vieta's formulas:
- Sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1 \)
- Product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-1}{1} = -1 \)

3. **Calculate \( \alpha^2 + \beta^2 \)**:
We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (1)^2 - 2(-1) = 1 + 2 = 3 \]

4. **Calculate \( \alpha^3 + \beta^3 \)**:
We can use the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]
We already know \( \alpha + \beta = 1 \), \( \alpha^2 + \beta^2 = 3 \), and \( \alpha \beta = -1 \).
Substituting these values:
\[ \alpha^3 + \beta^3 = 1 \left( 3 - (-1) \right) = 1 \times (3 + 1) = 1 \times 4 = 4 \]

### Final Answers:
- (i) \( \alpha^2 + \beta^2 = 3 \)
- (ii) \( \alpha^3 + \beta^3 = 4 \)