Find all integers $n>1$ such that any prime divisor of $n^6-1$ is a divisor of $(n^3-1)(n^2-1)$.
1. We start by considering the expression \( n^6 - 1 \). We can factorize it as follows:
   \[
   n^6 - 1 = (n^3 - 1)(n^3 + 1)
   \]
   Further factorizing each term, we get:
   \[
   n^3 - 1 = (n - 1)(n^2 + n + 1)
   \]
   \[
   n^3 + 1 = (n + 1)(n^2 - n + 1)
   \]
   Therefore,
   \[
   n^6 - 1 = (n - 1)(n^2 + n + 1)(n + 1)(n^2 - n + 1)
   \]

2. Next, we consider the expression \((n^3 - 1)(n^2 - 1)\). We can factorize it as follows:
   \[
   n^3 - 1 = (n - 1)(n^2 + n + 1)
   \]
   \[
   n^2 - 1 = (n - 1)(n + 1)
   \]
   Therefore,
   \[
   (n^3 - 1)(n^2 - 1) = (n - 1)(n^2 + n + 1)(n - 1)(n + 1) = (n - 1)^2(n + 1)(n^2 + n + 1)
   \]

3. We need to show that any prime divisor of \( n^6 - 1 \) is also a divisor of \((n^3 - 1)(n^2 - 1)\). This means we need to show:
   \[
   (n - 1)(n^2 + n + 1)(n + 1)(n^2 - n + 1) \text{ divides } (n - 1)^2(n + 1)(n^2 + n + 1)
   \]

4. Simplifying the divisibility condition, we see that:
   \[
   (n^2 - n + 1) \text{ must divide } (n - 1)
   \]
   This implies:
   \[
   \frac{n^2 - n + 1}{n - 1} \in \mathbb{Z}
   \]
   Simplifying the fraction:
   \[
   \frac{n^2 - n + 1}{n - 1} = \frac{n(n - 1) + 1}{n - 1} = n + \frac{1}{n - 1}
   \]
   For this to be an integer, \(\frac{1}{n - 1}\) must be an integer, which implies \(n - 1\) must be 1. Therefore, \(n = 2\).

5. We verify that \(n = 2\) works:
   \[
   n^6 - 1 = 2^6 - 1 = 63
   \]
   \[
   (n^3 - 1)(n^2 - 1) = (2^3 - 1)(2^2 - 1) = 7 \cdot 3 = 21
   \]
   The prime divisors of 63 are 3 and 7, and both are divisors of 21.

Thus, the only integer \(n > 1\) that satisfies the condition is \(n = 2\).

The final answer is \(\boxed{n = 2}\).