If $x^4 + x^2y^2 + y^4 = \frac{21}{256}$ and $x^2 + xy + y^2 = \frac{3}{16}$ then $2(x^2 + y^2) =$

$\frac{5}{8}$

If $x^4 + x^2y^2 + y^4 = \frac{21}{256}$

$x^2 + xy + y^2 = \frac{3}{16}$-----(A)

then $2(x^2 + y^2) =$ ?

x⁴ + x²y² + y⁴ = (x² – xy + y²) (x² + xy + y²)

comparing the equations we get,

$x^2 - xy + y^2 = \frac{21}{256} \times \frac{16}{3}  $ = $\frac{7}{16}$----(B)

From equations A and B =

2(x² + y²) = $\frac{7}{16}$ + $\frac{3}{16}$

so, 2(x² + y²) = ($\frac{7}{16}$ + $\frac{3}{16}$)

2(x² + y²) = (\(\frac{10}{16}\)) = \(\frac{5}{8}\)