If ` x + y = sqrt( 5) and x - y = sqrt(3)` ,t hen the value of ` 6 xy ( x^(2) + y^(2))` is

Options

A. 12
B. 10
C. 8
D. 6

Answer

To solve the problem, we need to find the value of \( 6xy(x^2 + y^2) \) given the equations \( x + y = \sqrt{5} \) and \( x - y = \sqrt{3} \).

### Step-by-Step Solution:

1. **Square both equations**:
- From the first equation \( x + y = \sqrt{5} \):
\[ (x + y)^2 = (\sqrt{5})^2 \implies x^2 + 2xy + y^2 = 5 \tag{1} \]
- From the second equation \( x - y = \sqrt{3} \):
\[ (x - y)^2 = (\sqrt{3})^2 \implies x^2 - 2xy + y^2 = 3 \tag{2} \]

2. **Add the two equations** (1) and (2):
\[ (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 5 + 3 \]
This simplifies to:
\[ 2x^2 + 2y^2 = 8 \implies x^2 + y^2 = 4 \tag{3} \]

3. **Substitute \( x^2 + y^2 \) back into one of the equations to find \( xy \)**:
- Substitute (3) into equation (1):
\[ 4 + 2xy = 5 \]
Rearranging gives:
\[ 2xy = 5 - 4 \implies 2xy = 1 \implies xy = \frac{1}{2} \tag{4} \]

4. **Now calculate \( 6xy(x^2 + y^2) \)**:
- Substitute \( xy \) from (4) and \( x^2 + y^2 \) from (3):
\[ 6xy(x^2 + y^2) = 6 \left(\frac{1}{2}\right)(4) \]
Simplifying this gives:
\[ = 6 \cdot \frac{1}{2} \cdot 4 = 3 \cdot 4 = 12 \]

Thus, the value of \( 6xy(x^2 + y^2) \) is \( \boxed{12} \).