What is the natural number n for which (3^9 + 3^{12} + 3^{15} + 3^n) is a perfect cube of an integer?

To find the natural number n for which the expression 3^9 + 3^{12} + 3^{15} + 3^n is a perfect cube, we first factor out the smallest common power, 3^9. This gives 3^9(1 + 3^3 + 3^6 + 3^{n-9}). Since 3^9 is already a perfect cube ((3^3)^3), the term in the bracket (1 + 27 + 729 + 3^{n-9}), which simplifies to 757 + 3^{n-9}, must also be a perfect cube. Testing the given options for n: if n=14, then n-9=5. The expression becomes 757 + 3^5 = 757 + 243 = 1000. Since 1000 is 10^3, it is a perfect cube. Thus, the entire expression 3^9 * 10^3 = (27 * 10)^3 = 270^3 is a perfect cube. Other options like n=13 result in 757 + 3^4 = 838, which is not a perfect cube. Therefore, n=14 is the correct solution.