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 equation \(x^3 + x + 1 = 0\).

### Step 1: Find the sums and products of the roots

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

- The sum of the roots \(\alpha + \beta + \gamma = 0\)
- 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 \(\alpha \beta \gamma = -1\)

### Step 2: Simplify the numerator

We can rewrite the numerator:

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

Using the fact that \(\alpha + \beta = -\gamma\), \(\beta + \gamma = -\alpha\), and \(\gamma + \alpha = -\beta\):

\[ = \alpha \beta (-\gamma) + \beta \gamma (-\alpha) + \gamma \alpha (-\beta) \]
\[ = -(\alpha \beta \gamma + \beta \gamma \alpha + \gamma \alpha \beta) \]
\[ = -3 \alpha \beta \gamma \]

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

\[ = -3(-1) = 3 \]

### Step 3: Simplify the denominator

Now we need to calculate \(\alpha^2 + \beta^2 + \gamma^2\). We can use the identity:

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

Substituting the values we found:

\[ = 0^2 - 2(1) = -2 \]

### Step 4: Combine the results

Now we can substitute the results back into our expression:

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

### Final Answer

Thus, the value of the expression is:

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