



 
Hi Arul. For a given positive number N to be a perfect square, it must satisfy S^{2} = N, where S is another positive number. Similarly, C^{3} = N for a perfect cube. N = S^{2} = C^{3} Thus, if N is a 6^{th} power, then this would certainly work. Say N = A^{6}. Then S = A^{3} and C = A^{2}. So pick 6^{th} powers. That's probably the easiest way to find solutions. Your two solutions are 64 = 2^{6} and 729 = 3^{6}. We can see that this is sufficient to determine it is a solution, but are all solutions 6^{th} powers? To answer this use the fact that every number is the unique product of a set of prime numbers (this is technically called the "Fundamental Theorem of Arithmetic"). Consider that N is a number that satisfies N = S^{2} = C^{3}. Then every prime factor that appears in the factorization of S appears twice as often in N, and there can be no other primes involved. So all N's prime factors appear an even number of times. Similarly, N's prime factors must appear an even "triple" of times to satisfy N = C^{3}. Thus each factor in N appears an even number of times and the number of times is a multiple of 3. So every factor appears 6, 12, 18, ... Hope this helps,  


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