Age puzzle across perfect powers: Ram's age was a perfect square last year and it will be a perfect cube next year. Starting from his current age, how many years must he wait until his age is again (after that) a perfect cube?

Introduction / Context:
This classic ages problem mixes number theory with linear reasoning. We are told that the present age x satisfies two near-neighbor conditions: last year (x − 1) was a perfect square, and next year (x + 1) will be a perfect cube. We then ask: counting forward from today, after next year's cube, how long until his age is once again a perfect cube?

Given Data / Assumptions:

• Current age: x (a positive integer).
• x − 1 = n^2 for some integer n ≥ 1.
• x + 1 = m^3 for some integer m ≥ 1.
• We seek the smallest t > 1 so that x + t is a perfect cube again (i.e., the next cube after x + 1).

Concept / Approach:
Subtract the two conditions: (x + 1) − (x − 1) = m^3 − n^2 ⇒ 2 = m^3 − n^2. Test small cubes until the right-hand side equals 2. With m = 3, m^3 = 27 and n^2 = 25, satisfying 27 − 25 = 2. Hence x + 1 = 27 ⇒ x = 26 and x − 1 = 25 = 5^2 (consistent).

Step-by-Step Solution:
1) Solve m^3 − n^2 = 2 ⇒ m = 3, n = 5.2) Current age x = 26; next year x + 1 = 27 = 3^3 (first cube).3) The next cube after 27 is 64 = 4^3.4) Waiting time from today to reach 64 is 64 − 26 = 38 years.

Verification / Alternative check:
Check neighbor conditions around x = 26: last year 25 is 5^2; next year 27 is 3^3. The next cube after 27 is indeed 64, so 38 years is correct.

Why Other Options Are Wrong:

• 39 years / 10 years / 64 years: Do not match the actual gap to the next cube after 27; 64 years is not a waiting time but a value.

Common Pitfalls:
Answering “1 year” by misreading “again” as the first cube (next year). The problem asks for the next cube after the immediate one.

Final Answer:
38 years