Let `alpha, beta and gamma` be the roots of equation `x^(3)+x+1=0`, then `(alpha beta(alpha+beta)+betagamma(beta+gamma)+gamma alpha(gamma+alpha))/(alpha^(2)+beta^(2)+gamma^(2))` is equal to

Options

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

Answer

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

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

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

### Step 1: Identify the coefficients of the polynomial
The polynomial can be compared with the standard form \(ax^3 + bx^2 + cx + d = 0\):
- Here, \(a = 1\), \(b = 0\), \(c = 1\), and \(d = 1\).

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- \(\alpha + \beta + \gamma = -\frac{b}{a} = 0\)
- \(\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = 1\)
- \(\alpha \beta \gamma = -\frac{d}{a} = -1\)

### Step 3: Simplify the numerator
We can express \(\alpha + \beta\), \(\beta + \gamma\), and \(\gamma + \alpha\) in terms of the roots:
- \(\alpha + \beta = -\gamma\)
- \(\beta + \gamma = -\alpha\)
- \(\gamma + \alpha = -\beta\)

Substituting these into the numerator:

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

Thus, the numerator becomes:

\[ -\alpha \beta \gamma - \beta \gamma \alpha - \gamma \alpha \beta = -3\alpha \beta \gamma \]

Since \(\alpha \beta \gamma = -1\):

\[ \text{Numerator} = -3(-1) = 3 \]

### Step 4: Simplify the denominator
To find \(\alpha^2 + \beta^2 + \gamma^2\), we can use the identity:

\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]

Substituting the known values:

\[ \alpha^2 + \beta^2 + \gamma^2 = 0^2 - 2(1) = -2 \]

### Step 5: Combine the results
Now we can substitute the numerator and denominator into the original expression:

\[ \frac{3}{-2} = -\frac{3}{2} \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{-\frac{3}{2}} \]