$\frac{a^3}{b^3}$+$\frac{a^3+1}{b^3+1}$+...+$\frac{a^3+2015}{b^3+2015}$=2016
b - positive integer, b can't be 0
a - real
Find $\frac{a^3}{b^3}$*$\frac{a^3+1}{b^3+1}$*...*$\frac{a^3+2015}{b^3+2015}$

1. Given the equation:
   \[
   \frac{a^3}{b^3} + \frac{a^3+1}{b^3+1} + \cdots + \frac{a^3+2015}{b^3+2015} = 2016
   \]
   where \( b \) is a positive integer and \( a \) is a real number.

2. We need to analyze the behavior of the terms \(\frac{a^3+n}{b^3+n}\) for \( n = 0, 1, 2, \ldots, 2015 \).

3. Consider the case \( a > b \):
   - For \( a > b \), \( a^3 > b^3 \).
   - Therefore, for each \( n \), \( a^3 + n > b^3 + n \).
   - This implies \(\frac{a^3+n}{b^3+n} > 1\) for all \( n = 0, 1, 2, \ldots, 2015 \).
   - Summing these terms would give:
     \[
     \frac{a^3}{b^3} + \frac{a^3+1}{b^3+1} + \cdots + \frac{a^3+2015}{b^3+2015} > 2016
     \]
   - This contradicts the given equation.

4. Consider the case \( a < b \):
   - For \( a < b \), \( a^3 < b^3 \).
   - Therefore, for each \( n \), \( a^3 + n < b^3 + n \).
   - This implies \(\frac{a^3+n}{b^3+n} < 1\) for all \( n = 0, 1, 2, \ldots, 2015 \).
   - Summing these terms would give:
     \[
     \frac{a^3}{b^3} + \frac{a^3+1}{b^3+1} + \cdots + \frac{a^3+2015}{b^3+2015} < 2016
     \]
   - This also contradicts the given equation.

5. The only remaining possibility is \( a = b \):
   - If \( a = b \), then for each \( n \):
     \[
     \frac{a^3+n}{b^3+n} = \frac{b^3+n}{b^3+n} = 1
     \]
   - Summing these terms would give:
     \[
     1 + 1 + \cdots + 1 = 2016
     \]
     which is consistent with the given equation.

6. Therefore, if \( a = b \), the product of the terms is:
   \[
   \frac{a^3}{b^3} \cdot \frac{a^3+1}{b^3+1} \cdots \frac{a^3+2015}{b^3+2015} = 1 \cdot 1 \cdot \cdots \cdot 1 = 1
   \]

The final answer is \(\boxed{1}\).