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.

Explore how to calculate the expression 10 times the sum of divisors of (n-1)^2, where n equals 3. Follow a step-by-step solution involving number theory and the sum of divisors function. Suitable for advanced high school students and anyone interested in number theory.

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