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 given problem, we need to find the equation whose roots are given in terms of the roots of the original cubic equation. Let's denote the roots of the original equation \( x^3 + ax^2 + b = 0 \) as \( \alpha_1, \alpha_2, \alpha_3 \).

### Step 1: Identify the roots of the new equation
The roots of the new equation are given as:
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. \( \frac{\alpha_1 \alpha_3 + \alpha_1 \alpha_2}{\alpha_1 \alpha_2 \alpha_3} \)

Let \( p = \alpha_1 \alpha_2 \alpha_3 \) (the product of the roots), \( q = \alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 \) (the sum of the products of the roots taken two at a time), and \( r = \alpha_1 + \alpha_2 + \alpha_3 \) (the sum of the roots).

### Step 2: Express the new roots in terms of \( p, q, r \)
The new roots can be rewritten as:
1. \( \frac{q - \alpha_3^2}{p} \)
2. \( \frac{q - \alpha_1^2}{p} \)
3. \( \frac{q - \alpha_2^2}{p} \)

### Step 3: Find the sum of the new roots
The sum of the new roots can be calculated as follows:
\[ \text{Sum} = \frac{(q - \alpha_3^2) + (q - \alpha_1^2) + (q - \alpha_2^2)}{p} = \frac{3q - (\alpha_1^2 + \alpha_2^2 + \alpha_3^2)}{p} \]
Using the identity \( \alpha_1^2 + \alpha_2^2 + \alpha_3^2 = r^2 - 2q \), we can substitute:
\[ \text{Sum} = \frac{3q - (r^2 - 2q)}{p} = \frac{5q - r^2}{p} \]

### Step 4: Find the sum of the products of the new roots taken two at a time
The sum of the products of the new roots taken two at a time is given by:
\[ \text{Product Sum} = \frac{(\frac{q - \alpha_3^2}{p})(\frac{q - \alpha_1^2}{p}) + (\frac{q - \alpha_2^2}{p})(\frac{q - \alpha_3^2}{p}) + (\frac{q - \alpha_1^2}{p})(\frac{q - \alpha_2^2}{p})}{p^2} \]
This simplifies to:
\[ \text{Product Sum} = \frac{(q^2 - q(\alpha_1^2 + \alpha_2^2 + \alpha_3^2) + \alpha_1^2 \alpha_2^2 + \alpha_2^2 \alpha_3^2 + \alpha_3^2 \alpha_1^2)}{p^2} \]
Using \( \alpha_1^2 + \alpha_2^2 + \alpha_3^2 = r^2 - 2q \) again, we can substitute:
\[ \text{Product Sum} = \frac{(q^2 - q(r^2 - 2q) + \text{(sum of squares of products)})}{p^2} \]

### Step 5: Find the product of the new roots
The product of the new roots is:
\[ \text{Product} = \frac{(q - \alpha_1^2)(q - \alpha_2^2)(q - \alpha_3^2)}{p^3} \]

### Step 6: Form the new polynomial
Using Vieta's formulas, we can form the new polynomial with roots as calculated above:
\[ x^3 - \left(\frac{5q - r^2}{p}\right)x^2 + \left(\text{Product Sum}\right)x - \left(\text{Product}\right) = 0 \]

### Final Equation
Thus, the final equation whose roots are the specified transformations of the original roots is:
\[ x^3 - \frac{5q - r^2}{p}x^2 + \left(\text{Product Sum}\right)x - \left(\text{Product}\right) = 0 \]