If $$x+y = 1$$, $$x^4+y^4 = 881$$, find the value of $$xy$$

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Uniteasy Admin · Accepted Answer

$$\begin{aligned} x^4+y^4  &= (x^2+y^2)^2-2x^2y^2 \  &=[(x+y)^2-2xy]^2-2x^2y^2 \ & = 881\end{aligned} $$c
Substitution of the given condition $$x+y = 1$$ results in a quadratic equation in terms of $$xy$$
$$(1-2xy)^2-2(xy)^2 = 881$$c
$$(xy)^2-2xy-440 = 0$$c
Factoring the LHS
$$(xy-22)(xy+20) = 0$$c
Therefore 
$$xy = 22 	ext{ or } xy = -20$$c

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