If `alpha,beta` are the roots of the equation `x^(2)+x+1=0`, find the value of `alpha^(3)-beta^(3)`.

Answer

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

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

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

### Step 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:
\[ \alpha^2 + \beta^2 = (-1)^2 - 2 \cdot 1 = 1 - 2 = -1 \]

### Step 4: Calculate \( \alpha - \beta \)
Using the identity:
\[ \alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} \]
Substituting the known values:
\[ \alpha - \beta = \sqrt{(-1)^2 - 4 \cdot 1} = \sqrt{1 - 4} = \sqrt{-3} = i\sqrt{3} \]

### Step 5: Calculate \( \alpha^3 - \beta^3 \)
Using the formula for the difference of cubes:
\[ \alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \beta^2 + \alpha\beta) \]
Substituting the values we have:
\[ \alpha^3 - \beta^3 = (i\sqrt{3})(-1 + 1) = (i\sqrt{3})(0) = 0 \]

### Final Answer
Thus, the value of \( \alpha^3 - \beta^3 \) is \( 0 \).

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