Find the equation whose roots are square of the roots of the equation `x^3-2x^2+3x+1=0`

Options

A. `x^3+2x^2+13x-1=0`
B. `x^3-13x^2+2x-1=0`
C. `x^3+3x^2-12x-1=0`
D. `-x^3+2x^2+13x+1=0`

Answer

To find the equation whose roots are the squares of the roots of the equation \( x^3 - 2x^2 + 3x + 1 = 0 \), we can follow these steps:

### Step 1: Identify the roots of the original equation
Let the roots of the equation \( x^3 - 2x^2 + 3x + 1 = 0 \) be \( \alpha, \beta, \gamma \).

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

### Step 3: Calculate the new roots
The new roots we need to find are \( \alpha^2, \beta^2, \gamma^2 \).

### Step 4: Find the sum of the new roots
The sum of the new roots is:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the values:
\[ \alpha^2 + \beta^2 + \gamma^2 = 2^2 - 2 \cdot 3 = 4 - 6 = -2 \]

### Step 5: 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:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the values:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = 3^2 - 2(-1)(2) = 9 + 4 = 13 \]

### Step 6: Find the product of the new roots
The product of the new roots is:
\[ \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-1)^2 = 1 \]

### Step 7: Form the new polynomial
Using the results from the previous steps, we can form the polynomial whose roots are \( \alpha^2, \beta^2, \gamma^2 \):
\[ x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots taken two at a time})x - (\text{product of roots}) = 0 \]
Substituting the values:
\[ x^3 - (-2)x^2 + 13x - 1 = 0 \]
This simplifies to:
\[ x^3 + 2x^2 + 13x - 1 = 0 \]

### Final Answer
The equation whose roots are the squares of the roots of the original equation is:
\[ \boxed{x^3 + 2x^2 + 13x - 1 = 0} \]