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

Options

A. 0
B. 3
C. -3
D. -1

Explanation

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

### Step 1: Identify the coefficients of the cubic equation
The given equation is:
\[ x^3 + 0x^2 + x + 1 = 0 \]
From this, we can identify:
- \( A = 1 \) (coefficient of \( x^3 \))
- \( B = 0 \) (coefficient of \( x^2 \))
- \( C = 1 \) (coefficient of \( x \))
- \( D = 1 \) (constant term)

### Step 2: Use Vieta's formulas to find the sums and products of the roots
According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{B}{A} = -\frac{0}{1} = 0 \)
- The sum of the product of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} = \frac{1}{1} = 1 \)
- The product of the roots \( \alpha\beta\gamma = -\frac{D}{A} = -\frac{1}{1} = -1 \)

### Step 3: Use the identity for the sum of cubes
There is an important identity that relates the sum of cubes of the roots to the roots themselves:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) \]
Since we know that \( \alpha + \beta + \gamma = 0 \), the second term becomes zero:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma \]

### Step 4: Substitute the product of the roots
Now, substituting the value of \( \alpha\beta\gamma \):
\[ \alpha^3 + \beta^3 + \gamma^3 = 3(-1) = -3 \]

### Final Answer
Thus, the value of \( \alpha^3 + \beta^3 + \gamma^3 \) is:
\[ \boxed{-3} \]