If `w`, `w^(2)` are the roots of `x^(2)+x+1=0` and `alpha, beta` are the roots of `x^(2)+px+q=0` then `(w alpha+w^(2)beta)(w^(2)alpha+w beta)` =

Options

A. `p^(2)-3q`
B. `p^(2)-2q`
C. p-q
D. `q^(2)-3p`

Answer

To solve the problem, we need to find the value of \((w \alpha + w^2 \beta)(w^2 \alpha + w \beta)\), given that \(w\) and \(w^2\) are the roots of the equation \(x^2 + x + 1 = 0\) and \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + px + q = 0\).

### Step 1: Identify the roots of the first equation
The roots of the equation \(x^2 + x + 1 = 0\) are given as \(w\) and \(w^2\).

Using Vieta's formulas:
- The sum of the roots \(w + w^2 = -\frac{b}{a} = -1\)
- The product of the roots \(w \cdot w^2 = \frac{c}{a} = 1\)

From the product of the roots, we can deduce that:
\[ w^3 = 1 \quad \text{(since \(w \cdot w^2 \cdot w = w^3\))} \]

### Step 2: Identify the roots of the second equation
For the second equation \(x^2 + px + q = 0\), we have:
- The sum of the roots \(\alpha + \beta = -p\)
- The product of the roots \(\alpha \cdot \beta = q\)

### Step 3: Expand the expression
We need to expand the expression \((w \alpha + w^2 \beta)(w^2 \alpha + w \beta)\).

Using the distributive property:
\[ (w \alpha + w^2 \beta)(w^2 \alpha + w \beta) = w \alpha \cdot w^2 \alpha + w \alpha \cdot w \beta + w^2 \beta \cdot w^2 \alpha + w^2 \beta \cdot w \beta \]

This simplifies to:
\[ = w^3 \alpha^2 + w^2 \alpha \beta + w^4 \beta \alpha + w^2 \beta^2 \]

### Step 4: Substitute \(w^3\) and \(w^4\)
Since \(w^3 = 1\) and \(w^4 = w\), we can substitute:
\[ = 1 \cdot \alpha^2 + w \cdot \alpha \beta + w^2 \cdot \alpha \beta + w^2 \beta^2 \]
\[ = \alpha^2 + (w + w^2) \alpha \beta + \beta^2 \]

### Step 5: Substitute \(w + w^2\)
From earlier, we know that \(w + w^2 = -1\):
\[ = \alpha^2 - \alpha \beta + \beta^2 \]

### Step 6: Use the identity for squares
Using the identity \(a^2 + b^2 = (a + b)^2 - 2ab\):
\[ = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting \(\alpha + \beta = -p\) and \(\alpha \beta = q\):
\[ = (-p)^2 - 2q = p^2 - 2q \]

### Step 7: Final expression
Thus, the value of \((w \alpha + w^2 \beta)(w^2 \alpha + w \beta)\) is:
\[ \boxed{p^2 - 3q} \]