$3 \cdot 6$ Find the smallest natural numbers $a$ and $b(b>1)$, such that $\sqrt{a \sqrt{a \sqrt{a}}}=b$.
(Kiev Mathematical Olympiad, 1978)
［Solution］From the problem, we have
$$
b^{8}=(\sqrt{a \sqrt{a \sqrt{a}}})^{8}=a^{7} \text {. }
$$

Let the prime number $p$ divide $b$. (Since $b>1$, such a prime number exists)
From (1), $p$ also divides $a$.
Let $b=p^{\beta} \cdot c, \quad a=p^{\alpha} \cdot d$, where $c$ and $d$ are natural numbers coprime with $p$, $\alpha \geqslant 1, \beta \geqslant 1$. Thus, we have
$$
p^{8 \beta} \cdot c^{8}=p^{7 \alpha} \cdot d^{7} \text {. }
$$

Since $\left(c^{8}, p\right)=1,\left(d^{7}, p\right)=1$, from (2) we have
$$
8 \beta=7 \alpha \text {. }
$$

Since $(7,8)=1$, then 7 divides $\beta$, and 8 divides $\alpha$, so we have
$$
\alpha \geqslant 8, \quad \beta \geqslant 7 \text {. }
$$

Since $c \geqslant 1, d \geqslant 1, \quad p \geqslant 2$.
Therefore, we must have
$$
\begin{array}{l}
a=p^{\alpha} \cdot d \geqslant p^{\alpha} \geqslant p^{8} \geqslant 2^{8}=256 \\
b=p^{\beta} \cdot c \geqslant p^{\beta} \geqslant p^{7} \geqslant 2^{7}=128 .
\end{array}
$$

Thus, $a=256, b=128$ are the smallest natural numbers that satisfy the given equation.