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 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 will use Vieta's formulas.

### Step 1: Identify coefficients
The general form of a cubic equation is given by:
\[ ax^3 + bx^2 + cx + d = 0 \]
From the given equation \(2x^3 + 3x^2 + 5x + 6 = 0\), we can identify:
- \(a = 2\)
- \(b = 3\)
- \(c = 5\)
- \(d = 6\)

### Step 2: Calculate \(\alpha + \beta + \gamma\)
According to Vieta's formulas, the sum of the roots \(\alpha + \beta + \gamma\) can be calculated as:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]
Substituting the values of \(b\) and \(a\):
\[ \alpha + \beta + \gamma = -\frac{3}{2} \]

### Step 3: Calculate \(\alpha\beta + \beta\gamma + \gamma\alpha\)
The sum of the products of the roots taken two at a time is given by:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \]
Substituting the values of \(c\) and \(a\):
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{5}{2} \]

### Step 4: Calculate \(\alpha\beta\gamma\)
The product of the roots is given by:
\[ \alpha\beta\gamma = -\frac{d}{a} \]
Substituting the values of \(d\) and \(a\):
\[ \alpha\beta\gamma = -\frac{6}{2} = -3 \]

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