If `alpha,beta,gamma` are the roots of the equation `(x^3+x^2+x+1)=0` then
`|{:(1+alpha," "1," "1),(" "1,1+beta," "1),(" "1," "1,1+gamma):}|` is equal to

Options

A. 1
B. 0
C. 2
D. none of these

Answer

To solve the given problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} 1 + \alpha & 1 & 1 \\ 1 & 1 + \beta & 1 \\ 1 & 1 & 1 + \gamma \end{vmatrix} \]

where \(\alpha, \beta, \gamma\) are the roots of the polynomial \(x^3 + x^2 + x + 1 = 0\).

### Step 1: Expand the Determinant

We will use the determinant expansion formula. The determinant can be expanded as follows:

\[ D = (1 + \alpha) \begin{vmatrix} 1 + \beta & 1 \\ 1 & 1 + \gamma \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 + \gamma \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 + \beta \\ 1 & 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

Now, we calculate the 2x2 determinants:

1. For the first determinant:

\[ \begin{vmatrix} 1 + \beta & 1 \\ 1 & 1 + \gamma \end{vmatrix} = (1 + \beta)(1 + \gamma) - (1)(1) = 1 + \beta + \gamma + \beta\gamma - 1 = \beta + \gamma + \beta\gamma \]

2. For the second determinant:

\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 + \gamma \end{vmatrix} = (1)(1 + \gamma) - (1)(1) = 1 + \gamma - 1 = \gamma \]

3. For the third determinant:

\[ \begin{vmatrix} 1 & 1 + \beta \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1 + \beta) = 1 - (1 + \beta) = -\beta \]

### Step 3: Substitute Back into the Determinant

Now substituting these back into the expression for \(D\):

\[ D = (1 + \alpha)(\beta + \gamma + \beta\gamma) - \gamma - \beta \]

### Step 4: Simplify the Expression

Expanding this gives:

\[ D = \beta + \gamma + \beta\gamma + \alpha(\beta + \gamma + \beta\gamma) - \gamma - \beta \]

Notice that \(\beta\) and \(\gamma\) cancel out:

\[ D = \alpha(\beta + \gamma + \beta\gamma) \]

### Step 5: Use Vieta's Formulas

From Vieta's formulas for the polynomial \(x^3 + x^2 + x + 1 = 0\):

- The sum of the roots \(\alpha + \beta + \gamma = -1\)
- The sum of the products of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = 1\)
- The product of the roots \(\alpha\beta\gamma = -1\)

Thus, we can express \(\beta + \gamma\) as:

\[ \beta + \gamma = -1 - \alpha \]

Substituting this into our expression for \(D\):

\[ D = \alpha(-1 - \alpha + \beta\gamma) \]

### Step 6: Find \(\beta\gamma\)

Using the product of the roots:

\[ \beta\gamma = \frac{-1}{\alpha} \]

Substituting this back into \(D\):

\[ D = \alpha(-1 - \alpha - \frac{1}{\alpha}) = \alpha(-1 - \alpha - \frac{1}{\alpha}) = \alpha(-\frac{\alpha + 1 + 1}{\alpha}) = -(\alpha + 1) \]

### Step 7: Evaluate \(D\)

Since \(\alpha, \beta, \gamma\) are the roots of the polynomial, substituting any of the roots into the determinant will yield \(0\) because the determinant is a linear combination of the roots.

Thus, we conclude:

\[ D = 0 \]

### Final Result

The value of the determinant is:

\[ \boxed{0} \]