If x is both the cube and the square of an integer and x is between 2

The question confused me a bit. Anyways, checking answer options is the best way to solve this.

(A) 8 ==> \(\sqrt{8}\) is not an integer. Incorrect.

(B) 16 ==> Cuberoot(16) is not an integer. Incorrect.

(C) 64 ==> Matches. Correct!

(D) 125 ==> \(\sqrt{125}\) is not an integer. Incorrect.

(E) 169 ==> Cuberoot(169) is not an integer. Incorrect.

clear 64
8^2 = 64
4^3 = 64

ans C

If a number is square of an integer, then the powers of all prime numbers in its prime factorisation Must be divisible by 2.
If a number is cube of an integer, then the powers of all prime numbers in its prime factorisation Must be divisible by 3.

Combining above two statements, it follows that if a number is square of an integer as well as cube of an integer, then powers of all prime numbers in its prime factorisation Must be Divisible by both 2 & 3 - or Divisible by 6.

Check options. A) 8 = 2^3. Powers are not multiples of 6, Eliminated
B) 16 = 2^4. Powers are not multiples of 6, Eliminated
C) 64 = 2^6. Here it is, powers are multiples of 6
D) 125 = 5^3. Powers are not multiples of 6, Eliminated
E) 169 = 13^2. Powers are not multiples of 6, Eliminated

Hence C answer

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