3. Solve the equation $x-\sqrt{x} \cdot 2^{-x^{6}}=2^{1-2 x^{6}}$.
3. Answer: $(6,12),(6,-12),(24,96)$.

Let Vasya perform operations with the number $x$, and Petya with the number $y$. Then $\left\{\begin{array}{l}2 x^{3}=3 y^{2}, \\ |x-y| \leqslant 100 .\end{array}\right.$ From the equation of the system, it follows that $x=3 a, y=2 b, a, b \in \mathbb{Z}$. Substituting, we get $54 a^{3}=12 b^{2} \Leftrightarrow 9 a^{3}=2 b^{2}$, hence $a=2 k, b=3 l, k, l \in \mathbb{Z}$, from which $4 k^{3}=l^{2}$. From this equation, it follows that $k=u^{2}, l=2 v, u, v \in \mathbb{Z}$, i.e., $u^{3}= \pm v$. The inequality of the system in terms of $k$ and $l$ will take the form $|6 k-6 l| \leqslant 100$, which, taking into account the integrality of $k$ and $l$, gives $|k-l| \leqslant 16$ or $\left|u^{2}-2 v\right| \leqslant 16$. From the inequality $\left|u^{2} \pm 2 u^{3}\right| \leqslant 16$ and the fact that $|1 \pm 2 u| \in \mathbb{Z}$ and $|1 \pm 2 u| \neq 0$, it follows that $|u|^{2} \leqslant 4 \Rightarrow|u| \leqslant 2$. From the equation of the system, it follows that $x \geqslant 0$, and therefore $u \geqslant 0$. Since $x$ and $y$ are different, $u$ is not equal to zero (otherwise $v$ is also zero). $x=6 u^{2}, y=12 v$. The remaining pairs are: $u=1, v= \pm 1 \Leftrightarrow(x, y)=(6, \pm 12) ; \quad u=2, v= \pm 8 \Leftrightarrow(x, y)=(24, \pm 96)$. The pairs $(x, y)=(6,12),(x, y)=(6,-12),(x, y)=(24,96)$ satisfy the inequality of the system.