Example 2 Given that $x$ and $y$ are real numbers, and satisfy $x y + x + y = 17, x^{2} y + x y^{2} = 66$.
Find the value of $x^{4} + x^{3} y + x^{2} y^{2} + x y^{3} + y^{4}$. (2000, Shandong Province Junior High School Mathematics Competition)
Explanation: According to the structural characteristics of the known equations, they can be transformed into
$$
x y+(x+y)=17, x y(x+y)=66 \text{. }
$$

Thus, a quadratic equation can be constructed with $x y$ and $x+y$ as its roots:
$$
t^{2}-17 t+66=0 \text{. }
$$

Solving this, we get $t_{1}=6, t_{2}=11$.
When $x y=6, x+y=11$, $x, y$ are the roots of the equation
$$
u^{2}-11 u+6=0
$$

Since $\Delta=(-11)^{2}-4 \times 6=121-24>0$, equation (1) has real roots.
$$
\begin{array}{l}
\text{Therefore, } x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4} \\
= x^{4}+x^{2} y^{2}+y^{4}+x y\left(x^{2}+y^{2}\right) \\
=\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}+x y\left(x^{2}+y^{2}\right) \\
= \left[(x+y)^{2}-2 x y\right]^{2}-(x y)^{2}+ \\
x y\left[(x+y)^{2}-2 x y\right] \\
= 12499 .
\end{array}
$$

When $x y=11, x+y=6$, $x, y$ are the roots of the equation
$$
u^{2}-6 u+11=0
$$

By inspection, the discriminant of equation (2) is less than 0, so equation (2) has no real roots.

In summary, $x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}=$ 12499.