If `alpha,beta` are the roots of the equation `x^(2)+px+q=0`, find the value of
(a) `alpha^(3)beta+alphabeta^(3)`
(b) `alpha^(4)+alpha^(2)beta^(2)+beta^(4)`.

Answer

To solve the problem step by step, we will find the values for both parts (a) and (b) as requested.

### Given:
The roots of the quadratic equation \( x^2 + px + q = 0 \) are \( \alpha \) and \( \beta \).

### Step 1: Find the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -p \)
- The product of the roots \( \alpha \beta = q \)

### Step 2: Solve part (a) \( \alpha^3 \beta + \alpha \beta^3 \)
We can factor this expression:
\[ \alpha^3 \beta + \alpha \beta^3 = \alpha \beta (\alpha^2 + \beta^2) \]
Now, we know \( \alpha \beta = q \). Next, we need to find \( \alpha^2 + \beta^2 \).

Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we have:
\[ \alpha^2 + \beta^2 = (-p)^2 - 2q = p^2 - 2q \]
Now substituting back into our expression for part (a):
\[ \alpha^3 \beta + \alpha \beta^3 = q(p^2 - 2q) \]
Thus, the final answer for part (a) is:
\[ \alpha^3 \beta + \alpha \beta^3 = qp^2 - 2q^2 \]

### Step 3: Solve part (b) \( \alpha^4 + \alpha^2 \beta^2 + \beta^4 \)
We can rewrite this expression as:
\[ \alpha^4 + \beta^4 + \alpha^2 \beta^2 \]
Using the identity for \( \alpha^4 + \beta^4 \):
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 \]
Substituting \( \alpha^2 + \beta^2 = p^2 - 2q \):
\[ \alpha^4 + \beta^4 = (p^2 - 2q)^2 - 2(\alpha \beta)^2 = (p^2 - 2q)^2 - 2q^2 \]
Now substituting back into our expression for part (b):
\[ \alpha^4 + \beta^4 + \alpha^2 \beta^2 = (p^2 - 2q)^2 - 2q^2 + q^2 \]
This simplifies to:
\[ (p^2 - 2q)^2 - q^2 \]
Now applying the identity \( a^2 - b^2 = (a - b)(a + b) \):
Let \( a = p^2 - 2q \) and \( b = q \):
\[ = (p^2 - 2q - q)(p^2 - 2q + q) = (p^2 - 3q)(p^2 - q) \]
Thus, the final answer for part (b) is:
\[ \alpha^4 + \alpha^2 \beta^2 + \beta^4 = (p^2 - 3q)(p^2 - q) \]

### Final Answers:
(a) \( \alpha^3 \beta + \alpha \beta^3 = qp^2 - 2q^2 \)

(b) \( \alpha^4 + \alpha^2 \beta^2 + \beta^4 = (p^2 - 3q)(p^2 - q) \)