4.99 Solve the system of equations $\left\{\begin{array}{l}x+y=4, \\ \left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=280 .\end{array}\right.$
(Kyiv Mathematical Olympiad, 1935)
[Solution] Substituting (1) into (2) yields
$$
\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}-x y\right)-70=0
$$

From (1), we get $x^{2}+y^{2}=16-2 x y$
Substituting (4) into (3) and simplifying, we get $3 x^{2} y^{2}-40 x y+93=0$,
so $x y=3$ or $x y=\frac{31}{3}$.
Combining with (1), we have
$$
\left\{\begin{array} { l } 
{ x + y = 4 , } \\
{ x y = 3 . }
\end{array} \text { or } \left\{\begin{array}{l}
x+y=4, \\
x y=\frac{31}{3} .
\end{array}\right.\right. \text { (no solution) }
$$
$\therefore$ The solution to the original system of equations is
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = 3 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
x_{2}=1, \\
y_{2}=3 .
\end{array}\right.\right.
$$