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}

Substitution of the given condition x+y = 1 results in a quadratic equation in terms of xy

(1-2xy)^2-2(xy)^2 = 881
(xy)^2-2xy-440 = 0

Factoring the LHS

(xy-22)(xy+20) = 0

Therefore

xy = 22 \text{ or } xy = -20


Steven Zheng posted 1 week ago

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