3. Find all pairs of natural numbers $a$ and $b$ such that $\left(a^{3}+b\right)\left(b^{3}+a\right)$ is a power of 2.
3. It is clear that $a$ and $b$ have the same parity. If $a=2^{r} a_{1}$ and $b=2^{s} b_{1}$, where $a_{1}$ and $b_{1}$ are odd and $0<r \leqslant s$, then $b^{3}+a=2^{r}\left(2^{3 s-r} b_{1}^{3}+a_{1}\right)$, so the odd number $2^{3 s-r} b_{1}^{3}+a_{1}$ is a power of two, which is impossible. Therefore, $a$ and $b$ are odd.

Assume that $a^{3}+b=2^{m}$ and $b^{3}+a=2^{n}$, where $m \leqslant n$. Then $a^{3}+b$ divides the numbers $a+b^{3}$ and $a^{9}+b^{3}$, hence $2^{m}=a^{3}+b \mid a^{9}-a=a\left(a^{2}-1\right)\left(a^{2}+1\right)\left(a^{4}+1\right)$. Since the numbers $a, \frac{a^{2}+1}{2}$ and $\frac{a^{4}+1}{2}$ are odd, it follows that $a^{3}+b \mid 4\left(a^{2}-1\right)$. From this, we have $a^{3}<4\left(a^{2}-1\right)$, i.e., $a<4$.

If $a=1$, then $b+1$ and $b^{3}+1$ are powers of two, so is the number $\frac{b^{3}+1}{b+1}=b^{2}-b+1$, which is odd, so it must be $b^{2}-b+1=1$ and $b=1$.

If $a=3$, then $b+3^{3}$ divides $4\left(3^{2}-1\right)=32$, so we get $b=5$.

Therefore, the only solutions $(a, b)$ are $(1,1),(3,5)$ and $(5,3)$.