3. If real numbers $x, y$ satisfy
$$
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \frac{x}{5^{3}+4^{2}}+\frac{y}{5^{3}+6^{3}}=1,
$$

then $x+y=$ $\qquad$ .
3.432 .

It is known that $3^{3}$ and $5^{3}$ are the two roots of the equation $\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1$, which means they are the roots of the 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
$$
\begin{array}{l}
3^{3}+5^{3}=-\left(4^{3}+6^{3}-x-y\right) . \\
\text { Therefore, } x+y=3^{3}+4^{3}+5^{3}+6^{3}=432 \text {. }
\end{array}
$$