If `alpha, beta ,gamma ` are the roots of the equation `x ^(3) + 2x ^(2) - x+1 =0,` then vlaue of `((2- alpha )(2-beta) (2-gamma))/((2+ alpha ) (2+ beta ) (2 + gamma))` is :

Options

A. 5
B. `-5`
C. 10
D. `5/3`

Answer

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

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

where \(\alpha, \beta, \gamma\) are the roots of the cubic equation:

\[ x^3 + 2x^2 - x + 1 = 0 \]

### Step 1: Identify the coefficients of the cubic equation

From the given equation, we have:
- \(a = 1\)
- \(b = 2\)
- \(c = -1\)
- \(d = 1\)

### Step 2: Use Vieta's formulas to find the sums and products of the roots

According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{2}{1} = -2\)
- The sum of the products of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-1}{1} = -1\)
- The product of the roots \(\alpha\beta\gamma = -\frac{d}{a} = -\frac{1}{1} = -1\)

### Step 3: Expand the numerator and denominator

Now we will expand both the numerator and the denominator.

**Numerator:**

\[ (2 - \alpha)(2 - \beta)(2 - \gamma) = 2^3 - 2^2(\alpha + \beta + \gamma) + 2(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma \]

Substituting the values from Vieta's formulas:

\[ = 8 - 4(-2) + 2(-1) - (-1) \]
\[ = 8 + 8 - 2 + 1 = 15 \]

**Denominator:**

\[ (2 + \alpha)(2 + \beta)(2 + \gamma) = 2^3 + 2^2(\alpha + \beta + \gamma) + 2(\alpha\beta + \beta\gamma + \gamma\alpha) + \alpha\beta\gamma \]

Substituting the values:

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

### Step 4: Combine the results

Now we can substitute the results of the numerator and denominator back into the expression:

\[ \frac{(2 - \alpha)(2 - \beta)(2 - \gamma)}{(2 + \alpha)(2 + \beta)(2 + \gamma)} = \frac{15}{-3} = -5 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{-5} \]