Let `f(x)=x^(3)+x+1`, let `p(x)` be a cubic polynomial such that the roots of `p(x)=0` are the squares of the roots of `f(x)=0` , then

Options

A. `p(1)=3`
B. the value of `P(n)`, `n in N` is odd
C. Sum of all roots of `p(x)=0` is `-2`
D. Sum of all product of roots taken two at a time is `1`

Explanation

To solve the problem, we need to find the polynomial \( p(x) \) whose roots are the squares of the roots of the polynomial \( f(x) = x^3 + x + 1 \).

### Step-by-step Solution:

1. **Identify the Roots of \( f(x) \)**:
Let the roots of \( f(x) = 0 \) be \( \alpha_1, \alpha_2, \alpha_3 \). According to Vieta's formulas, we have:
- Sum of roots: \( \alpha_1 + \alpha_2 + \alpha_3 = 0 \)
- Sum of products of roots taken two at a time: \( \alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1 = 1 \)
- Product of roots: \( \alpha_1\alpha_2\alpha_3 = -1 \)

2. **Roots of \( p(x) \)**:
The roots of \( p(x) \) are \( \alpha_1^2, \alpha_2^2, \alpha_3^2 \). We need to find the polynomial whose roots are these squares.

3. **Sum of the Roots of \( p(x) \)**:
The sum of the roots \( \alpha_1^2 + \alpha_2^2 + \alpha_3^2 \) can be calculated using the identity:
\[ \alpha_1^2 + \alpha_2^2 + \alpha_3^2 = (\alpha_1 + \alpha_2 + \alpha_3)^2 - 2(\alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1) \]
Substituting the values from Vieta's:
\[ \alpha_1^2 + \alpha_2^2 + \alpha_3^2 = 0^2 - 2 \cdot 1 = -2 \]

4. **Sum of the Products of Roots Taken Two at a Time**:
The sum of the products of the roots taken two at a time \( \alpha_1^2\alpha_2^2 + \alpha_2^2\alpha_3^2 + \alpha_3^2\alpha_1^2 \) can be calculated as:
\[ \alpha_1^2\alpha_2^2 + \alpha_2^2\alpha_3^2 + \alpha_3^2\alpha_1^2 = (\alpha_1\alpha_2)^2 + (\alpha_2\alpha_3)^2 + (\alpha_3\alpha_1)^2 \]
Using the identity:
\[ (\alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1)^2 = (\alpha_1\alpha_2)^2 + (\alpha_2\alpha_3)^2 + (\alpha_3\alpha_1)^2 + 2\alpha_1\alpha_2\alpha_3(\alpha_1 + \alpha_2 + \alpha_3) \]
Substituting the known values:
\[ 1^2 = (\alpha_1\alpha_2)^2 + (\alpha_2\alpha_3)^2 + (\alpha_3\alpha_1)^2 + 2(-1)(0) \]
Thus,
\[ (\alpha_1\alpha_2)^2 + (\alpha_2\alpha_3)^2 + (\alpha_3\alpha_1)^2 = 1 \]

5. **Product of the Roots**:
The product of the roots \( \alpha_1^2\alpha_2^2\alpha_3^2 \) is given by:
\[ (\alpha_1\alpha_2\alpha_3)^2 = (-1)^2 = 1 \]

6. **Construct the Polynomial \( p(x) \)**:
Using the results from the previous steps, we can construct the polynomial \( p(x) \) using Vieta's relations:
\[ p(x) = x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots taken two at a time})x - (\text{product of roots}) \]
Substituting the values:
\[ p(x) = x^3 - (-2)x^2 + 1x - 1 = x^3 + 2x^2 + x - 1 \]

### Final Result:
The polynomial \( p(x) \) is:
\[ p(x) = x^3 + 2x^2 + x - 1 \]