If a, b are the real roots of `x^(2) + px + 1 = 0` and c, d are the real roots of `x^(2) + qx + 1 = 0`, then `(a-c)(b-c)(a+d)(b+d)` is divisible by

Options

A. `a - b - c - d`
B. `a + b + c - d`
C. `a + b + c + d`
D. `a - b - c - d`

Explanation

To solve the problem, we need to analyze the two quadratic equations given and the expression we need to evaluate for divisibility.

### Step-by-Step Solution:

1. **Identify the Quadratic Equations**:
The first quadratic equation is given as:
\[ x^2 + px + 1 = 0 \]
The roots of this equation are \( a \) and \( b \).

The second quadratic equation is:
\[ x^2 + qx + 1 = 0 \]
The roots of this equation are \( c \) and \( d \).

2. **Use Vieta's Formulas**:
From Vieta's formulas, we know:
- For the first equation:
- Sum of roots: \( a + b = -p \)
- Product of roots: \( ab = 1 \)

- For the second equation:
- Sum of roots: \( c + d = -q \)
- Product of roots: \( cd = 1 \)

3. **Rewrite the Expression**:
We need to evaluate the expression:
\[ (a - c)(b - c)(a + d)(b + d) \]
We can rearrange this as:
\[ (a - c)(b + d)(b - c)(a + d) \]

4. **Expand the Expression**:
Let's expand the expression:
\[ (a - c)(b - c) = ab - ac - bc + c^2 \]
\[ (a + d)(b + d) = ab + ad + bd + d^2 \]

Therefore, the full expression becomes:
\[ (ab - ac - bc + c^2)(ab + ad + bd + d^2) \]

5. **Substituting Known Values**:
Since \( ab = 1 \) and \( cd = 1 \), we can substitute these values into the expression:
\[ (1 - ac - bc + c^2)(1 + ad + bd + d^2) \]

6. **Simplify the Expression**:
Now we simplify:
\[ = (1 - ac - bc + c^2)(1 + ad + bd + d^2) \]
This will yield a more complex expression, but we are interested in its divisibility.

7. **Final Form**:
The expression simplifies to:
\[ c^2 + d^2 + 2 - (a^2 + b^2 + 2) \]
which can be expressed as:
\[ (c + d)^2 - (a + b)^2 \]
Using the identity \( x^2 - y^2 = (x - y)(x + y) \):
\[ = (c + d - (a + b))(c + d + (a + b)) \]

8. **Conclusion**:
Therefore, the expression \( (a - c)(b - c)(a + d)(b + d) \) is divisible by:
\[ (c + d - (a + b))(c + d + (a + b)) \]