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

$$x^4 +x^2y^2+y^4 = \dfrac {21} {256} $$ and $$x^2 + xy + y^2 = \dfrac{3}{16}$$ than $$2(x^2 +y^2)= ?$$

$$x^4 + x^2y^2 + y^4   $$................ euestion (1)

$$\Rightarrow x^4 +2x^2 y^2 +y^4 -x^2y^2  $$(add & substract x^2 y^2 in equestion (1))

$$\Rightarrow(x^2 +y^2)^2- (xy)^2$$

$$\Rightarrow (x^2 +y^2 +xy)(x^2 +y^2 -xy) = \dfrac{21}{256}  we   use   that   formula {a^2 -b^2 = (a+b)(a-b)}$$

$$\Rightarrow \dfrac{3}{16} (x^2 +y^2 -xy) =\dfrac {7}{16} $$

$$\Rightarrow (x^2+y^2-xy)=\dfrac {7}{16} $$ ............equestion (a)

$$\Rightarrow (x^2 +y^2 +xy) = \dfrac {3}{16}$$ ...........equestion (b)

Adding equestion (a) + equestion (b)

$$\Rightarrow 2(x^2 +y^2) = \dfrac {10}{16}$$

$$\Rightarrow 2(x^2 +y^2 ) = \dfrac {5}{8}$$ Ans .