Answer

Given

x^4+x^2=3
(1)
y^4-y^2=3
(2)

Subtract (2) from (1)

x^4+x^2-(y^4-y^2) =0

Group terms for factorization

x^4-y^4+(x^2+y^2) = 0

(x^2+y^2)(x^2-y^2+1)=0

From given conditions (1), (2), x,y\ne 0

Therefore

x^2-y^2+1 = 0

x^2-y^2 = -1
(3)

Addition of (1) and (2) gives the equation

x^4+x^2+y^4-y^2 = 6

Then

x^4+y^4 = 6-(x^2-y^2)

Substitute value of x^2-y^2 given in (3)

x^4+y^4 = 6-(-1) = 7

Steven Zheng posted 4 days ago

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