If $$x^{4}+ x^{2} y^{2}+y^{4}=21$$, and $$x^{2}+xy+y^{2} = 3,$$ then what is the value of 4xy?

Given that $$x^{4}+ x^{2} y^{2}+y^{4}=21$$

 $$x^{2}+xy+y^{2} = 3,$$  Equestion (1)

$$\Rightarrow (x^{2} + xy + y^{2}) ^ {2} = (3)^2 $$ (Sqaring both side (1))

$$\Rightarrow x^{4} +x^{2}y^{2} + y^{4} + 2 x^{3}y +2xy^{3} + 2 x^{2}y^{2} = 9 $$ 

(From the formula$$ (a+b+c)^2 = a^2 +b^2 +c^2  + 2ab+2bc+2ca $$)

$$\Rightarrow 21 + 2 x^{3}y + 2xy^{3}+2x^2y^2 = 9 $$ (put the value from given question )

$$\Rightarrow  2 x^{3}y + 2xy^{3}+2x^2y^2 =9-21$$

$$\Rightarrow 2 x^{3}y + 2xy^{3}+2x^2y^2 = -12 $$

$$\Rightarrow 2xy (x^2 +y^2 + xy) = -12 $$ (taken a common 2xy)

$$\Rightarrow 2xy (3) = -12 $$ (put the value from equestion (1))

$$\Rightarrow 2xy = \dfrac {-12}{3} =  - 4 $$

$$\Rightarrow 4xy = - 8 $$ Ans (multiply by 2 both side )