If `alpha,beta,gamma ` are the roots of
`4x^3-6x^2+7x+3=0 ` then find the value of `alpha beta +beta gamma+gamma alpha`.

Answer

To find the value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) for the roots \( \alpha, \beta, \gamma \) of the polynomial equation \( 4x^3 - 6x^2 + 7x + 3 = 0 \), we can use Vieta's formulas.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given polynomial is \( 4x^3 - 6x^2 + 7x + 3 = 0 \). We can identify the coefficients:
- \( A = 4 \) (coefficient of \( x^3 \))
- \( B = -6 \) (coefficient of \( x^2 \))
- \( C = 7 \) (coefficient of \( x \))
- \( D = 3 \) (constant term)

2. **Apply Vieta's Formulas**:
According to Vieta's formulas for a cubic equation \( Ax^3 + Bx^2 + Cx + D = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{B}{A} \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} \)
- The product of the roots \( \alpha\beta\gamma = -\frac{D}{A} \)

3. **Calculate \( \alpha\beta + \beta\gamma + \gamma\alpha \)**:
From Vieta's formulas, we have:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} \]
Substituting the values of \( C \) and \( A \):
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{7}{4} \]

4. **Final Result**:
Therefore, the value of \( \alpha \beta + \beta \gamma + \gamma \alpha \) is \( \frac{7}{4} \).