If ` alpha , beta , gamma ` are the roots of the equation `x^3 +4x^2 -5x +3=0` then ` sum (1)/( alpha^2 beta^2)=`

Options

A. `-9/2`
B. `-(38 )/( 49 )`
C. `(26 )/(9)`
D. `98`

Answer

To find the value of \( \sum \frac{1}{\alpha^2 \beta^2} \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + 4x^2 - 5x + 3 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is \( x^3 + 4x^2 - 5x + 3 = 0 \). We can identify the coefficients:
- \( a = 1 \)
- \( b = 4 \)
- \( c = -5 \)
- \( d = 3 \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas, we have:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{4}{1} = -4 \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -5 \)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{3}{1} = -3 \)

### Step 3: Find \( \sum \frac{1}{\alpha^2 \beta^2} \)
We can express \( \sum \frac{1}{\alpha^2 \beta^2} \) as follows:
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\gamma^2 \alpha^2} \]
This can be rewritten using the product of the roots:
\[ = \frac{\gamma^2 + \alpha^2 + \beta^2}{(\alpha\beta\gamma)^2} \]
Substituting \( \alpha\beta\gamma = -3 \):
\[ = \frac{\gamma^2 + \alpha^2 + \beta^2}{(-3)^2} = \frac{\gamma^2 + \alpha^2 + \beta^2}{9} \]

### Step 4: Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the values we found:
\[ = (-4)^2 - 2(-5) = 16 + 10 = 26 \]

### Step 5: Substitute back into the equation
Now we can substitute \( \alpha^2 + \beta^2 + \gamma^2 \) back into our expression for \( \sum \frac{1}{\alpha^2 \beta^2} \):
\[ \sum \frac{1}{\alpha^2 \beta^2} = \frac{26}{9} \]

### Final Answer
Thus, the value of \( \sum \frac{1}{\alpha^2 \beta^2} \) is \( \frac{26}{9} \).

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