If ` alpha , beta , gamma ` are the roots of ` x^3 +2x^2 +3x +8=0` then ` (3-alpha )(3 - beta) (3-gamma)`=

Options

A. 52
B. 62
C. 65
D. 67

Answer

To solve the problem, we need to find the value of \( (3 - \alpha)(3 - \beta)(3 - \gamma) \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + 2x^2 + 3x + 8 = 0 \).

### Step 1: Identify the coefficients
The polynomial can be expressed in the standard form \( ax^3 + bx^2 + cx + d = 0 \). Here, we have:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 3 \)
- \( d = 8 \)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{2}{1} = -2 \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{3}{1} = 3 \)
- \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{8}{1} = -8 \)

### Step 3: Expand \( (3 - \alpha)(3 - \beta)(3 - \gamma) \)
We can expand this expression:
\[ (3 - \alpha)(3 - \beta)(3 - \gamma) = 3^3 - 3^2(\alpha + \beta + \gamma) + 3(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma \]

### Step 4: Substitute the values from Vieta's
Now, substituting the values we found using Vieta's:
- \( 3^3 = 27 \)
- \( 3^2(\alpha + \beta + \gamma) = 9(-2) = -18 \)
- \( 3(\alpha\beta + \beta\gamma + \gamma\alpha) = 3(3) = 9 \)
- \( -\alpha\beta\gamma = -(-8) = 8 \)

### Step 5: Combine the results
Now, we combine all these results:
\[ (3 - \alpha)(3 - \beta)(3 - \gamma) = 27 - (-18) + 9 + 8 \]
\[ = 27 + 18 + 9 + 8 \]
\[ = 62 \]

### Final Answer
Thus, the value of \( (3 - \alpha)(3 - \beta)(3 - \gamma) \) is \( 62 \).

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