Let `alpha,beta,gamma` be the roots of `x^(3)+2x^(2)-x-3=0`. If the absolute value of the expression `(alpha-1)/(alpha+2)+(beta-1)/(beta+2)+(gamma-1)/(gamma+2)` can be expressed as `(m)/(n)` where `m` and `n` are co-prime the value of `|{:(m,n^(2)),(m-n,m+n):}|` is

Options

A. 17
B. 27
C. 37
D. 47

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{\alpha - 1}{\alpha + 2} + \frac{\beta - 1}{\beta + 2} + \frac{\gamma - 1}{\gamma + 2} \]

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

### Step 1: Identify the coefficients and use Vieta's formulas

From the polynomial \(x^3 + 2x^2 - x - 3 = 0\), we can identify:
- \(a = 1\)
- \(b = 2\)
- \(c = -1\)
- \(d = -3\)

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

### Step 2: Simplify the expression

The expression can be rewritten as:

\[ \frac{\alpha - 1}{\alpha + 2} = 1 - \frac{3}{\alpha + 2} \]

Thus, we can express the entire sum as:

\[ \left(1 - \frac{3}{\alpha + 2}\right) + \left(1 - \frac{3}{\beta + 2}\right) + \left(1 - \frac{3}{\gamma + 2}\right) \]

This simplifies to:

\[ 3 - 3\left(\frac{1}{\alpha + 2} + \frac{1}{\beta + 2} + \frac{1}{\gamma + 2}\right) \]

### Step 3: Find \(\frac{1}{\alpha + 2} + \frac{1}{\beta + 2} + \frac{1}{\gamma + 2}\)

Using the identity for the sum of reciprocals:

\[ \frac{1}{\alpha + 2} + \frac{1}{\beta + 2} + \frac{1}{\gamma + 2} = \frac{(\beta + 2)(\gamma + 2) + (\gamma + 2)(\alpha + 2) + (\alpha + 2)(\beta + 2)}{(\alpha + 2)(\beta + 2)(\gamma + 2)} \]

Calculating the numerator:

\[ (\beta + 2)(\gamma + 2) + (\gamma + 2)(\alpha + 2) + (\alpha + 2)(\beta + 2) = \beta\gamma + 2\beta + 2\gamma + 4 + \gamma\alpha + 2\gamma + 2\alpha + 4 + \alpha\beta + 2\alpha + 2\beta + 4 \]

This simplifies to:

\[ (\alpha\beta + \beta\gamma + \gamma\alpha) + 2(\alpha + \beta + \gamma) + 12 = -1 + 2(-2) + 12 = -1 - 4 + 12 = 7 \]

Calculating the denominator:

\[ (\alpha + 2)(\beta + 2)(\gamma + 2) = \alpha\beta\gamma + 2(\alpha\beta + \beta\gamma + \gamma\alpha) + 4(\alpha + \beta + \gamma) + 8 \]

This simplifies to:

\[ 3 + 2(-1) + 4(-2) + 8 = 3 - 2 - 8 + 8 = 1 \]

Thus, we have:

\[ \frac{1}{\alpha + 2} + \frac{1}{\beta + 2} + \frac{1}{\gamma + 2} = \frac{7}{1} = 7 \]

### Step 4: Substitute back into the expression

Now substituting back:

\[ 3 - 3 \cdot 7 = 3 - 21 = -18 \]

### Step 5: Find the absolute value

The absolute value of the expression is:

\[ |-18| = 18 \]

### Step 6: Express as \(\frac{m}{n}\)

We can express \(18\) as \(\frac{18}{1}\), where \(m = 18\) and \(n = 1\). Since \(m\) and \(n\) are co-prime, we can proceed to calculate the determinant.

### Step 7: Calculate the determinant

We need to find the value of the determinant:

\[ \left| \begin{array}{cc} m & n^2 \\ m - n & m + n \end{array} \right| \]

Substituting \(m = 18\) and \(n = 1\):

\[ \left| \begin{array}{cc} 18 & 1 \\ 17 & 19 \end{array} \right| = (18)(19) - (1)(17) = 342 - 17 = 325 \]

### Final Answer

Thus, the final value of the determinant is:

\[ \boxed{325} \]