If $$x,y$$ are real numbers such that $$x^4+x^2=3,y^4-y^2=3$$．
Find the value of $$x^4+y^4$$．

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Uniteasy Admin · Accepted Answer

Given
$$x^4+x^2=3$$n
$$y^4-y^2=3$$n
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), $$x,y
e 0 $$
Therefore
$$x^2-y^2+1 = 0$$
$$x^2-y^2 = -1$$n
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)
$$x^4+y^4 = 6-(-1) = 7$$

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