If ` x^(2) - xy + y^(2) = 2 and x^(4) + x^(2) y^(2) + y^(4) = 6` then the value of `( x ^(2) + xy+ y^(2))` is

Options

A. 1
B. 12
C. 3
D. 36

Answer

To solve the problem, we need to find the value of \( x^2 + xy + y^2 \) given the equations:

1. \( x^2 - xy + y^2 = 2 \) (Equation 1)
2. \( x^4 + x^2y^2 + y^4 = 6 \) (Equation 2)

### Step 1: Rewrite Equation 1
From Equation 1, we can express \( x^2 + y^2 \) in terms of \( xy \):
\[ x^2 + y^2 = 2 + xy \]

### Step 2: Substitute into Equation 2
Now, we need to express \( x^4 + y^4 \) in terms of \( x^2 + y^2 \) and \( xy \). We know that:
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]
Substituting \( x^2 + y^2 \) from Step 1:
\[ x^4 + y^4 = (2 + xy)^2 - 2x^2y^2 \]
Expanding this:
\[ x^4 + y^4 = 4 + 4xy + x^2y^2 - 2x^2y^2 = 4 + 4xy - x^2y^2 \]

### Step 3: Substitute into Equation 2
Now, substituting \( x^4 + y^4 \) into Equation 2:
\[ (4 + 4xy - x^2y^2) + x^2y^2 = 6 \]
This simplifies to:
\[ 4 + 4xy = 6 \]
Subtracting 4 from both sides:
\[ 4xy = 2 \]
Dividing by 4:
\[ xy = \frac{1}{2} \]

### Step 4: Substitute \( xy \) back into Equation 1
Now we can substitute \( xy = \frac{1}{2} \) back into the expression we found for \( x^2 + y^2 \):
\[ x^2 + y^2 = 2 + \frac{1}{2} = \frac{5}{2} \]

### Step 5: Calculate \( x^2 + xy + y^2 \)
Now we can find \( x^2 + xy + y^2 \):
\[ x^2 + xy + y^2 = (x^2 + y^2) + xy = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3 \]

Thus, the value of \( x^2 + xy + y^2 \) is \( \boxed{3} \).