Solve the following system of equations:

$$
\begin{aligned}
x^{3} y+x y^{3} & =10 \\
x^{4}+y^{4} & =17
\end{aligned}
$$
Solution. With the substitutions $x y=a, x^{2}+y^{2}=b$, the equations can be written as $a b=10, b^{2}-2 a^{2}=17$. Expressing the value of $a$ from the first equation and substituting it into the second equation, we get

$$
b^{2}-2(10 / b)^{2}-17=0, \quad \text { that is } \quad b^{4}-17 b^{2}-200=0
$$

Solving the quadratic equation for $b^{2}$, we find that $b^{2}=25$, the other root being negative. Since $b \geq 0$, we get $b=5, a=2$. From this, we find that $(x+y)^{2}=\left(x^{2}+y^{2}\right)+2 x y=b+2 a=9$,

$$
(x-y)^{2}=\left(x^{2}+y^{2}\right)-2 x y=b-2 a=1
$$

thus $x+y= \pm 3, x-y= \pm 1$. We arrive at four possibilities according to the four sign combinations:

$$
x_{1}=2, y_{1}=1 ; \quad x_{2}=1, \quad y_{2}=2 ; \quad x_{3}=-1, y_{3}=-2 ; \quad x_{4}=-2, y_{4}=-1
$$

The system of equations is satisfied by all four pairs of numbers.