3. Find all solutions of the system of equations: $x+y=4,\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=280$.
9.3. Answer. (3,1) and $(1,3)$.

From the condition $x+y=4$ - not equal to 0, so in the second equation, we can cancel $x+y=4$, obtaining $\left(x^{2}+y^{2}\right)\left(x^{2}-x y+y^{2}\right)=70$. Further, squaring the first equation, we get $x^{2}+y^{2}=16-2 x y$. Substituting into the second, we get $3(x y)^{2}-40 x y+93=0$, from which $x y=3$ or $x y=\frac{31}{3}$. Together with the first equation, we get a system for $x, y$. In the first case, $x=3, y=1$ or $x=1, y=3$, in the second case there are no solutions.

Grading Remarks. If both answers are guessed and verified - 1 point is given. If there is a correct solution with the acquisition of any extraneous roots, 3 points are deducted. If one solution is lost in the process, 3 points are deducted.