If `x+y=sqrt3, x-y=sqrt2`, then the expression `8xy(x^2+y^2)` has the value

Options

A. `5sqrt2`
B. `10sqrt2`
C. 20
D. 5

Answer

To solve the problem, we need to find the value of the expression \( 8xy(x^2 + y^2) \) given the equations \( x + y = \sqrt{3} \) and \( x - y = \sqrt{2} \).

### Step 1: Square the equations
First, we will square both equations to find \( x^2 + y^2 \) and \( xy \).

1. From \( x + y = \sqrt{3} \):
\[ (x + y)^2 = (\sqrt{3})^2 \]
\[ x^2 + 2xy + y^2 = 3 \quad \text{(Equation 1)} \]

2. From \( x - y = \sqrt{2} \):
\[ (x - y)^2 = (\sqrt{2})^2 \]
\[ x^2 - 2xy + y^2 = 2 \quad \text{(Equation 2)} \]

### Step 2: Add the equations
Next, we will add Equation 1 and Equation 2 to eliminate \( xy \):

\[ (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 3 + 2 \]
\[ 2x^2 + 2y^2 = 5 \]
\[ x^2 + y^2 = \frac{5}{2} \quad \text{(Equation 3)} \]

### Step 3: Substitute to find \( xy \)
Now, we will substitute \( x^2 + y^2 \) back into one of the original equations to find \( xy \). We will use Equation 1:

\[ x^2 + 2xy + y^2 = 3 \]
Substituting \( x^2 + y^2 = \frac{5}{2} \):
\[ \frac{5}{2} + 2xy = 3 \]
Subtract \( \frac{5}{2} \) from both sides:
\[ 2xy = 3 - \frac{5}{2} \]
Convert \( 3 \) to a fraction:
\[ 2xy = \frac{6}{2} - \frac{5}{2} = \frac{1}{2} \]
Thus,
\[ xy = \frac{1}{4} \]

### Step 4: Calculate the expression
Now we can calculate \( 8xy(x^2 + y^2) \):
\[ 8xy(x^2 + y^2) = 8 \cdot \frac{1}{4} \cdot \frac{5}{2} \]
Calculating this step-by-step:
1. \( 8 \cdot \frac{1}{4} = 2 \)
2. \( 2 \cdot \frac{5}{2} = 5 \)

Thus, the value of \( 8xy(x^2 + y^2) \) is \( 5 \).

### Final Answer
The value of the expression \( 8xy(x^2 + y^2) \) is \( \boxed{5} \).