If `2x^(3) + 3x^(2) + 5x +6=0` has roots `alpha, beta, gamma` then find `alpha + beta + gamma, alphabeta + betagamma + gammaalpha` and `alpha beta gamma`

Answer

To solve the given cubic equation \( 2x^3 + 3x^2 + 5x + 6 = 0 \) and find the values of \( \alpha + \beta + \gamma \), \( \alpha\beta + \beta\gamma + \gamma\alpha \), and \( \alpha\beta\gamma \), we can use Vieta's formulas.

### Step-by-Step Solution:

1. **Identify coefficients**:
The given equation is in the form \( ax^3 + bx^2 + cx + d = 0 \).
Here, we have:
- \( a = 2 \)
- \( b = 3 \)
- \( c = 5 \)
- \( d = 6 \)

2. **Sum of the roots** \( \alpha + \beta + \gamma \):
According to Vieta's formulas, the sum of the roots of the polynomial is given by:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]
Substituting the values:
\[ \alpha + \beta + \gamma = -\frac{3}{2} \]

3. **Sum of the products of the roots taken two at a time** \( \alpha\beta + \beta\gamma + \gamma\alpha \):
This is given by:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \]
Substituting the values:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{5}{2} \]

4. **Product of the roots** \( \alpha\beta\gamma \):
This is given by:
\[ \alpha\beta\gamma = -\frac{d}{a} \]
Substituting the values:
\[ \alpha\beta\gamma = -\frac{6}{2} = -3 \]

### Final Answers:
- \( \alpha + \beta + \gamma = -\frac{3}{2} \)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{5}{2} \)
- \( \alpha\beta\gamma = -3 \)