If the roots of equation `x^(3) + ax^(2) + b = 0 are alpha _(1), alpha_(2), and `
` alpha_(3) (a , b ne 0)`. Then find the equation whose roots are
`(alpha_(1)alpha_(2)+alpha_(2)alpha_(3))/(alpha_(1)alpha_(2)alpha_(3)), (alpha_(2)alpha_(3)+alpha_(3)alpha_(1))/(alpha_(1)alpha_(2)alpha_(3)), (alpha_(1)alpha_(3)+alpha_(1)alpha_(2))/(alpha_(1)alpha_(2)alpha_(3)) `.

Explanation

To solve the problem, we need to find the equation whose roots are given by:

1. \(\frac{\alpha_1 \alpha_2 + \alpha_2 \alpha_3}{\alpha_1 \alpha_2 \alpha_3}\)
2. \(\frac{\alpha_2 \alpha_3 + \alpha_3 \alpha_1}{\alpha_1 \alpha_2 \alpha_3}\)
3. \(\frac{\alpha_1 \alpha_3 + \alpha_1 \alpha_2}{\alpha_1 \alpha_2 \alpha_3}\)

where \(\alpha_1, \alpha_2, \alpha_3\) are the roots of the equation \(x^3 + ax^2 + b = 0\).

### Step 1: Identify the relationships among the roots
From Vieta's formulas for the polynomial \(x^3 + ax^2 + b = 0\), we know:
- \(\alpha_1 + \alpha_2 + \alpha_3 = -a\)
- \(\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = 0\) (since the coefficient of \(x\) is 0)
- \(\alpha_1 \alpha_2 \alpha_3 = -b\)

### Step 2: Calculate the new roots
Using the relationships from Vieta's formulas, we can simplify the new roots:

1. For the first root:
\[ \frac{\alpha_1 \alpha_2 + \alpha_2 \alpha_3}{\alpha_1 \alpha_2 \alpha_3} = \frac{-\alpha_3}{-b} = \frac{\alpha_3}{b} \]

2. For the second root:
\[ \frac{\alpha_2 \alpha_3 + \alpha_3 \alpha_1}{\alpha_1 \alpha_2 \alpha_3} = \frac{-\alpha_1}{-b} = \frac{\alpha_1}{b} \]

3. For the third root:
\[ \frac{\alpha_1 \alpha_3 + \alpha_1 \alpha_2}{\alpha_1 \alpha_2 \alpha_3} = \frac{-\alpha_2}{-b} = \frac{\alpha_2}{b} \]

Thus, the new roots can be expressed as:
- \(r_1 = \frac{\alpha_1}{b}\)
- \(r_2 = \frac{\alpha_2}{b}\)
- \(r_3 = \frac{\alpha_3}{b}\)

### Step 3: Form the new polynomial
The new polynomial whose roots are \(r_1, r_2, r_3\) can be found using Vieta's relations again:
- The sum of the roots:
\[ r_1 + r_2 + r_3 = \frac{\alpha_1 + \alpha_2 + \alpha_3}{b} = \frac{-a}{b} \]

- The product of the roots:
\[ r_1 r_2 r_3 = \frac{\alpha_1 \alpha_2 \alpha_3}{b^3} = \frac{-b}{b^3} = -\frac{1}{b^2} \]

- The sum of the products of the roots taken two at a time:
\[ r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1}{b^2} = \frac{0}{b^2} = 0 \]

### Step 4: Write the polynomial
Using the above results, the polynomial can be expressed as:
\[ x^3 - \left(\frac{-a}{b}\right)x^2 + 0 \cdot x - \left(-\frac{1}{b^2}\right) = 0 \]
This simplifies to:
\[ x^3 + \frac{a}{b}x^2 + \frac{1}{b^2} = 0 \]

Multiplying through by \(b^2\) to eliminate the fractions gives:
\[ b^2 x^3 + ab x^2 + 1 = 0 \]

### Final Answer
The required equation whose roots are \(\frac{\alpha_1 \alpha_2 + \alpha_2 \alpha_3}{\alpha_1 \alpha_2 \alpha_3}, \frac{\alpha_2 \alpha_3 + \alpha_3 \alpha_1}{\alpha_1 \alpha_2 \alpha_3}, \frac{\alpha_1 \alpha_3 + \alpha_1 \alpha_2}{\alpha_1 \alpha_2 \alpha_3}\) is:
\[ b^2 x^3 + ab x^2 + 1 = 0 \]