Example 3 If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \\
\frac{x}{5^{3}+4^{3}}+\frac{y}{5^{3}+6^{3}}=1,
\end{array}
$$

then $x+y=$ $\qquad$
(2005, National Junior High School Mathematics Competition)
【Analysis】It is easy to notice that the denominators of the two given equations contain $3^{3}$ and $5^{3}$ respectively. Therefore, we can consider $3^{3}$ and $5^{3}$ as the roots of a certain equation.
Solution From the given conditions, it is easy to see that $3^{3}$ and $5^{3}$ are the roots of the equation in terms of $t$:
$$
\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1 \text {, }
$$

which is equivalent to the quadratic equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0.
$$
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
3^{3}+5^{3}=-\left(4^{3}+6^{3}-x-y\right) . \\
\text { Therefore, } x+y=3^{3}+4^{3}+5^{3}+6^{3}=432 .
\end{array}
$$