If $x^{4}+ x^{2} y^{2}+y^{4}=21$, and $x^{2}+xy+y^{2} = 3,$ then what is the value of 4xy?
Answer & explanation
Correct answer: option 4
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