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

Let's denote \( a = x^2 + y^2 \) and \( b = xy \). We can rewrite the first equation as:

\[ a - b = 2 \quad \text{(Equation 1)} \]

From this, we can express \( a \) in terms of \( b \):

\[ a = b + 2 \]

Next, we can rewrite the second equation using the identities involving \( a \) and \( b \):

\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = a^2 - 2b^2 \]

Thus, we can rewrite the second equation as:

\[ a^2 - 2b^2 + b^2 = 6 \]

This simplifies to:

\[ a^2 - b^2 = 6 \quad \text{(Equation 2)} \]

Now we have two equations:

1. \( a - b = 2 \)
2. \( a^2 - b^2 = 6 \)

From Equation 1, we can substitute \( a \) in Equation 2:

Substituting \( a = b + 2 \) into Equation 2:

\[ (b + 2)^2 - b^2 = 6 \]

Expanding this gives:

\[ b^2 + 4b + 4 - b^2 = 6 \]

This simplifies to:

\[ 4b + 4 = 6 \]

Subtracting 4 from both sides:

\[ 4b = 2 \]

Dividing by 4:

\[ b = \frac{1}{2} \]

Now substituting \( b \) back into the expression for \( a \):

\[ a = b + 2 = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]

Finally, we need to find \( x^2 + xy + y^2 \):

\[ x^2 + xy + y^2 = a + b = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3 \]

Thus, the value of \( x^2 + xy + y^2 \) is:

\[ \boxed{3} \]