If `alpha, beta , gamma` are the zeros of the polynomial `x^(3)-6x^(2)-x+30` then the value of `(alpha beta+beta gamma+gamma alpha)` is

Options

A. `-1 `
B. 1
C. `-5`
D. 30

Answer

To find the value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) for the polynomial \( x^3 - 6x^2 - x + 30 \), we can use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial:**
The given polynomial is \( x^3 - 6x^2 - x + 30 \). We can identify the coefficients as follows:
- Coefficient of \( x^3 \) (let's call it \( a \)): \( 1 \)
- Coefficient of \( x^2 \) (let's call it \( b \)): \( -6 \)
- Coefficient of \( x \) (let's call it \( c \)): \( -1 \)
- Constant term (let's call it \( d \)): \( 30 \)

2. **Apply Vieta's Formulas:**
According to Vieta's formulas for a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-6}{1} = 6 \)
- 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 \( \alpha \beta \gamma = -\frac{d}{a} = -\frac{30}{1} = -30 \)

3. **Conclusion:**
Thus, the value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) is \( -1 \).

### Final Answer:
The value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) is \( -1 \).