This suggests that $ f(n) = n^3 $ for $ n = 3, 4, 5, 6 $. However, $ f(x) $ is a cubic polynomial, and $ x^3 $ is already a cubic polynomial. Since a cubic is uniquely determined by four points, and $ f(n) = n^3 $ satisfies all four conditions, we conclude:

Thu, 30 Apr 2026 02:37:06 +0000

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