To solve the expression $10 \cdot \sigma((n-1)^2)$ where $n = 3$:

1. First, substitute $n = 3$ into the expression $(n-1)^2$: $(n-1)^2 = (3-1)^2 = 2^2 = 4$

2. Next, find the value of $\sigma(4)$, where $\sigma(m)$ represents the sum of the divisors of $m$.

   The divisors of 4 are 1, 2, and 4. Therefore: $\sigma(4) = 1 + 2 + 4 = 7$

3. Finally, multiply this result by 10: $10 \cdot 7 = 70$

So, the final answer is $70$.

Would you like to delve into any specific steps, or do you have further questions?

Here are 5 related questions to consider:

1. What is the sum of the divisors function $\sigma(m)$ for a prime number?
2. How do you compute $\sigma(n)$ for a general positive integer $n$?
3. What is the value of $10 \cdot \sigma((n-1)^2)$ when $n = 4$?
4. Can $\sigma(n)$ ever be greater than $n^2$?
5. How does the divisor sum function $\sigma(n)$ relate to perfect numbers?

Tip: Remember that the divisor sum function $\sigma(n)$ plays a key role in number theory, especially in the study of perfect and abundant numbers.