If one root of `x^(2) + px + 1 = 0` is the cube of the other root then p =

Options

A. 0
B. 1
C. `1, +-2`
D. `0, +-2`

Answer

To solve the problem, we need to find the value of \( p \) given that one root of the quadratic equation \( x^2 + px + 1 = 0 \) is the cube of the other root.

Let's denote the roots of the equation as \( \alpha \) and \( \beta \). According to the problem, we can assume:

\[ \beta = \alpha^3 \]

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know that:
- The sum of the roots \( \alpha + \beta = -p \)
- The product of the roots \( \alpha \cdot \beta = 1 \)

Substituting \( \beta = \alpha^3 \) into these formulas:

1. **Sum of the roots:**
\[ \alpha + \alpha^3 = -p \]

2. **Product of the roots:**
\[ \alpha \cdot \alpha^3 = \alpha^4 = 1 \]

### Step 2: Solve for \( \alpha \)
From the product of the roots equation \( \alpha^4 = 1 \), we can find the possible values of \( \alpha \):
\[ \alpha = 1 \quad \text{or} \quad \alpha = -1 \]

### Step 3: Find corresponding \( p \) values
Now we will find \( p \) for both values of \( \alpha \).

#### Case 1: \( \alpha = 1 \)
If \( \alpha = 1 \):
\[ \beta = \alpha^3 = 1^3 = 1 \]
Now substituting into the sum of roots:
\[ 1 + 1 = -p \implies 2 = -p \implies p = -2 \]

#### Case 2: \( \alpha = -1 \)
If \( \alpha = -1 \):
\[ \beta = \alpha^3 = (-1)^3 = -1 \]
Now substituting into the sum of roots:
\[ -1 - 1 = -p \implies -2 = -p \implies p = 2 \]

### Step 4: Conclusion
We have found two possible values for \( p \):
- \( p = -2 \)
- \( p = 2 \)

### Final Answer
Thus, the values of \( p \) are \( p = -2 \) and \( p = 2 \).