If `alpha, beta and gamma` are the roots of the equation `x^(3)-3x^(2)+4x+4=0`, then the value of `|(a^(2)+1,1,1),(1,beta^(2)+1,1),(1,1,gamma^(2)+1)|` is equal to

Options

A. 32
B. 16
C. 56
D. 64

Answer

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

\[ D = \begin{vmatrix} \alpha^2 + 1 & 1 & 1 \\ 1 & \beta^2 + 1 & 1 \\ 1 & 1 & \gamma^2 + 1 \end{vmatrix} \]

where \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the polynomial \(x^3 - 3x^2 + 4x + 4 = 0\).

### Step 1: Identify the coefficients and calculate the roots' properties
From the polynomial \(x^3 - 3x^2 + 4x + 4 = 0\), we can identify the coefficients:
- \(a = 1\)
- \(b = -3\)
- \(c = 4\)
- \(d = 4\)

Using Vieta's formulas, we can find:
1. \(\alpha + \beta + \gamma = -\frac{b}{a} = 3\)
2. \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 4\)
3. \(\alpha\beta\gamma = -\frac{d}{a} = -4\)

### Step 2: Simplify the determinant
We can simplify the determinant \(D\) using properties of determinants. We will perform row operations to make the calculations easier.

1. Subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]

This gives us:
\[ D = \begin{vmatrix} \alpha^2 + 1 & 1 & 1 \\ 1 - (\alpha^2 + 1) & \beta^2 + 1 - 1 & 0 \\ 1 - (\alpha^2 + 1) & 0 & \gamma^2 + 1 - 1 \end{vmatrix} \]
which simplifies to:
\[ D = \begin{vmatrix} \alpha^2 + 1 & 1 & 1 \\ -\alpha^2 & \beta^2 & 0 \\ -\alpha^2 & 0 & \gamma^2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we can calculate the determinant using the formula for a \(3 \times 3\) determinant:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Here, \(a = \alpha^2 + 1\), \(b = 1\), and \(c = 1\), and the other elements are:
- \(d = -\alpha^2\)
- \(e = \beta^2\)
- \(f = 0\)
- \(g = -\alpha^2\)
- \(h = 0\)
- \(i = \gamma^2\)

Calculating \(D\):
\[ D = (\alpha^2 + 1)(\beta^2 \cdot 0 - 0 \cdot \gamma^2) - 1(-\alpha^2 \cdot 0 - 0 \cdot -\alpha^2) + 1(-\alpha^2 \cdot 0 - \beta^2 \cdot -\alpha^2) \]
This simplifies to:
\[ D = 0 + 0 + \alpha^2 \beta^2 \]

### Step 4: Evaluate the determinant
Since we have \(\alpha^2 \beta^2\) and we know from Vieta's relations that \(\alpha, \beta, \gamma\) are roots of the polynomial, we can conclude that the determinant simplifies to zero due to the linear dependence of the rows.

Thus, the value of the determinant is:
\[ \boxed{0} \]