3. Given $\sqrt{x}+\sqrt{y}=35, \sqrt[3]{x}+\sqrt[3]{y}=13$. Then $x+y=$ $\qquad$
3.793 .
$$
\begin{aligned}
x+y= & (\sqrt[3]{x}+\sqrt[3]{y})\left[(\sqrt[3]{x}+\sqrt[3]{y})^{2}-3 \sqrt[3]{x y}\right] \\
= & 13\left[13^{2}-3 \sqrt[3]{x y}\right], \\
& (\sqrt{x}+\sqrt{y})^{2}=x+y+2 \sqrt{x y}=35^{2}, \\
& \therefore 35^{2}-2 \sqrt{x y}=13^{3}-39 \sqrt[3]{x y},
\end{aligned}
$$

i.e., $2 \sqrt{x y}-39 \sqrt[3]{x y}+972=0$.
Let $\sqrt[6]{x y}=t$, the above equation becomes
$$
2 t^{3}-39 t^{2}+972=0,
$$

i.e.,
$$
\begin{array}{l}
2 t^{3}-12 t^{2}-27 t^{2}+972=0, \\
2 t^{2}(t-6)-27(t-6)(t+6)=0, \\
(t-6)(t-18)(2 t+9)=0 . \\
t_{1}=6, t_{2}=18, t_{3}=-\frac{9}{2} \text { (discard). }
\end{array}
$$

When $t_{2}=18$, $\sqrt[3]{x y}=324$, at this time $x+y=13\left(13^{2}\right.$ $-3 \times 324)$ is a negative value, which does not meet the problem's requirements;
When $t_{1}=6$, $\sqrt[6]{x y}=6^{6}, \sqrt[3]{x y}=36$, $x+y=13\left(13^{2}-3 \times 36\right)=793$.