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

Answers (1)

x^2+xy+y^2=3
By Squaring both sides,we get
x^4+x^2y^2+y^4+2x^3y+2xy^3+2x^2y^2=9
21+2xy(x^2+xy+y^2)=9
21+2xy(3)=9
6xy=-12
xy=-2
4xy=-8

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