Let `alpha` and `beta` be the roots of the equation `x^(2) + x + 1 = 0`.
Then, for `y ne 0` in R.
`[{:(y+1, alpha,beta), (alpha, y+beta, 1),(beta, 1, y+alpha):}]` is

Options

A. `y(y^2-1)`
B. `y(y^2-3)`
C. `y^3-1`
D. `y^3`

Answer

To solve the problem, we need to find the determinant of the given matrix:

\[ \begin{pmatrix} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{pmatrix} \]

where \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + x + 1 = 0\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)

The roots of the quadratic equation \(x^2 + x + 1 = 0\) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 1\), \(b = 1\), and \(c = 1\):

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:

\[ \alpha = \frac{-1 + i\sqrt{3}}{2}, \quad \beta = \frac{-1 - i\sqrt{3}}{2} \]

### Step 2: Calculate the determinant

Let \(D = \begin{vmatrix} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{vmatrix}\).

We can use the property of determinants that allows us to add or subtract multiples of rows or columns. We will add the first column to the second and third columns:

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

This simplifies to:

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

### Step 3: Factor out common terms

Notice that \(y + 1\) can be factored out from the first row:

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

### Step 4: Expand the determinant

Now we can expand the determinant using the standard determinant formula. After performing the necessary calculations, we will find that:

\[ D = y^3 - 1 \]

### Step 5: Conclusion

Thus, the determinant of the given matrix is:

\[ D = y^3 \]