Example 6 Consider the following action as one operation: cutting a rope of positive integer length into three ropes of positive integer length, with the longest rope being exactly one. Given a rope of length $a^{b}$ (where $a$ and $b$ are both integers greater than 1), two players, A and B, take turns performing the following operation: A starts by performing one operation on the rope, then B performs one operation on the longest rope resulting from A's operation, ... Each player performs an operation on the longest rope among the three ropes resulting from the previous operation by the other player. If a player cannot perform an operation when it is their turn, the other player wins. Who has a winning strategy?
【Analysis】Obviously, if $1 \leqslant n \leqslant 3$, A must fail; if $4 \leqslant n \leqslant 7=3+2+2$, then A has a winning strategy; if $8 \leqslant n \leqslant 9$, then the length of the longest rope after A's operation must satisfy $4 \leqslant m \leqslant 7$, so B has a winning strategy; if $10 \leqslant n \leqslant 25=9+8+8$, then A has a winning strategy; if $26 \leqslant n \leqslant 27$, then the length of the longest rope after A's operation must satisfy $10 \leqslant m \leqslant 25$, so B has a winning strategy. Using mathematical induction, we can prove:
B has a winning strategy $\Leftrightarrow n=3^{k}-1,3^{k}$.
If $a$ is a power of 3, then $a^{b}$ is also a power of 3, and B has a winning strategy.
Below, let's assume $a$ is not a power of 3, and
$$
a^{b}=3^{k}-1(k \geqslant 2) \text {. }
$$

Then $a^{b} \equiv 2(\bmod 3) \Rightarrow b$ is odd
$$
\begin{array}{l}
\Rightarrow 3^{k}=a^{b}+1=(a+1)\left(a^{b-1}-a^{b-2}+\cdots-a+1\right) \\
\Rightarrow a+1=3^{i} \Rightarrow a=3^{i}-1 \\
\Rightarrow a \equiv-1(\bmod 3) \\
\Rightarrow 3^{k-i}=a^{b-1}-a^{b-2}+\cdots-a+1 \equiv b(\bmod 3)
\end{array}
$$
$\Rightarrow b$ is a multiple of 3.
Let $c=a^{\frac{b}{3}}$. Then
$$
\begin{aligned}
3^{k} & =c^{3}+1=(c+1)\left((c+1)^{2}-3 c\right) \\
\Rightarrow & \left(c+1,(c+1)^{2}-3 c\right) \\
& =(c+1,3 c)=(c+1,3)=3,
\end{aligned}
$$

and $c+1, (c+1)^{2}-3 c$ are both powers of 3, which can only be
$$
\begin{array}{l}
(c+1)^{2}-3 c=3 \text { or } c+1=3 \\
\Rightarrow c=2, k=2, b=3, a=2 .
\end{array}
$$

In summary, when $a=2, b=3$ or $a$ is a power of 3, B has a winning strategy; otherwise, A has a winning strategy.