2. If real numbers $x, y$ satisfy
$$
\begin{array}{l}
\frac{x}{2^{10}+5^{3}}+\frac{y}{2^{10}+6^{3}}=1, \\
\frac{x}{3^{10}+5^{3}}+\frac{y}{3^{10}+6^{3}}=1,
\end{array}
$$

then $x+y=$ $\qquad$ (the result should be written in the form of a power).
From the given, we know that $2^{10}$ and $3^{10}$ are the two roots of the equation $\frac{x}{t+5^{3}}+\frac{y}{t+6^{3}}=1$ with respect to $t$, which means they are the roots of
$$
t^{2}-\left(x+y-5^{3}-6^{3}\right) t+\left(5^{3} \times 6^{3}-5^{3} y-6^{3} x\right)=0
$$

Therefore, $2^{10}+3^{10}=x+y-5^{3}-6^{3}$.
$$
\text { Hence } x+y=2^{10}+3^{10}+5^{3}+6^{3} \text {. }
$$