If`alpha , beta , gamma` are the roots of ` x^3 + px^2 + qx + r=0` form the equation whose roots are ` alpha beta , beta gamma , gamma alpha `

Answer

To find the equation whose roots are \( \alpha \beta, \beta \gamma, \gamma \alpha \) given that \( \alpha, \beta, \gamma \) are the roots of the cubic equation \( x^3 + px^2 + qx + r = 0 \), we can follow these steps:

### Step 1: Identify the relationships from the original equation
From the cubic equation \( x^3 + px^2 + qx + r = 0 \), we can derive the following relationships based on Vieta's formulas:
- The sum of the roots \( \alpha + \beta + \gamma = -p \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
- The product of the roots \( \alpha\beta\gamma = -r \)

### Step 2: Find the sum of the new roots
The new roots are \( \alpha\beta, \beta\gamma, \gamma\alpha \). The sum of these roots can be calculated as follows:
\[ \text{Sum} = \alpha\beta + \beta\gamma + \gamma\alpha = q \]

### Step 3: Find the sum of the products of the new roots taken two at a time
Next, we calculate the sum of the products of the new roots taken two at a time:
\[ \text{Sum of products} = \alpha\beta \cdot \beta\gamma + \beta\gamma \cdot \gamma\alpha + \gamma\alpha \cdot \alpha\beta \]
This can be simplified as:
\[ = \alpha\beta\gamma(\alpha + \beta + \gamma) = (-r)(-p) = pr \]

### Step 4: Find the product of the new roots
Now, we find the product of the new roots:
\[ \text{Product} = \alpha\beta \cdot \beta\gamma \cdot \gamma\alpha = (\alpha\beta\gamma)^2 = (-r)^2 = r^2 \]

### Step 5: Form the new polynomial
Using the relationships derived, we can now write the new polynomial whose roots are \( \alpha\beta, \beta\gamma, \gamma\alpha \):
\[ x^3 - (\text{Sum of roots}) x^2 + (\text{Sum of products}) x - (\text{Product}) = 0 \]
Substituting the values we found:
\[ x^3 - qx^2 + prx - r^2 = 0 \]

### Final Answer
Thus, the equation whose roots are \( \alpha\beta, \beta\gamma, \gamma\alpha \) is:
\[ x^3 - qx^2 + prx - r^2 = 0 \]

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