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

4 -4 12 -8

-8

x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2) If $x^{4}+ x^{2} y^{2}+y^{4}=21$, $x^{2}+xy+y^{2} = 3$-----(A) Then, $x^{2}-xy+y^{2} = \frac{21}{3}=7$------(B) From equations A and B we get, 2xy = -4 xy = -2 4xy = -8