If `alpha ,beta and gamma` are the roots of
`x^3-2x^2+3x-4 =0 `, then find
`sumalpha^2beta^2`

Answer

To find the value of \( \sum \alpha^2 \beta^2 \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial equation \( x^3 - 2x^2 + 3x - 4 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is:
\[ x^3 - 2x^2 + 3x - 4 = 0 \]
From this, we can identify the coefficients:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 3 \)
- \( d = -4 \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for the roots \( \alpha, \beta, \gamma \):
1. The sum of the roots:
\[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-2}{1} = 2 \]
2. The sum of the products of the roots taken two at a time:
\[ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = \frac{3}{1} = 3 \]
3. The product of the roots:
\[ \alpha \beta \gamma = -\frac{d}{a} = -\frac{-4}{1} = 4 \]

### Step 3: Find \( \sum \alpha^2 \beta^2 \)
We want to find:
\[ \sum \alpha^2 \beta^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 \]
This can be rewritten using the identity:
\[ \sum \alpha^2 \beta^2 = (\alpha \beta + \beta \gamma + \gamma \alpha)^2 - 2 \alpha \beta \gamma (\alpha + \beta + \gamma) \]

### Step 4: Substitute the known values
Substituting the values we found:
\[ \sum \alpha^2 \beta^2 = (3)^2 - 2 \cdot 4 \cdot 2 \]
Calculating this gives:
\[ \sum \alpha^2 \beta^2 = 9 - 16 = -7 \]

### Final Answer
Thus, the value of \( \sum \alpha^2 \beta^2 \) is:
\[ \boxed{-7} \]