Question

x^4+x^2y^2+y^4=21 and x^2+xy+y^2=7 than find value of 1/x^2+1/y^2​

Answer 1

Answer:

x^4+x^2y^2+y^4+^x^2y^2=21+x^2y^2

(x^2+y^2)^2=21+x^y^2

x^2+y^2=7-xy

(7-xy)^2=21+x^2y^2

49+x^2y^2-14xy=21+x^2y^2

-14xy=-28

xy=2

(1/x+1/y)^2=(1/x)^2+(1/y)2+2/(xy)

x^2+y^2=7-2=5

(x^2+y^2)/xy=5/xy=5/2

(1/x+1/y)=5/2

(5/2)^2=(1/x)^2+(1/y)^2+1

(25/4)-1= 1/x^2+1/y^2=21/4Ans