8
81
1
64 8*8=64 2^6=64
No, 2 is neither a perfect square nor a perfect cube.
128 is neither a perfect square nor a perfect cube, it is 2⁷ (2 to the 7th power).
To be a perfect square, a number must have a square root that evaluates to an integer. The square root of 2 is approximately equal to 1.414, thus it is not a perfect square.
To be a perfect square, all the primes in a number's prime factorisation must have an even power To be a perfect cube, all the primes in a number's prime factorisation must a power that is a multiple of 3 → To be a perfect square, all the primes in a number's prime factorisation must a power that is a multiple of 3 and a multiple of 2, ie the power must be a multiple of 6 The smallest prime is 2 2⁶ = 64 = (2³)² = 8² = (2²)³ = 4³ 2¹² = 4096 (too large) 3⁶ = 729 (too large) There is also 1 = 1² = 1³ Thus the whole numbers less than 100 which are both perfect squares and perfect cubes are 1 and 64.
The idea is to take out perfect squares. The largest perfect square in this case is 256, which is the square of 16 (if you have trouble figuring this out, you can take out a smaller perfect square first, and then see if you find additional perfect squares). In any case, the end result should not have a factor that is a perfect square. Using the symbol "root()" for square root: root(512) = root(256 x 2) = root(256) x root(2) = 16 root(2)
No. Though every perfect square is a rational number, not every rational number is a perfect square. Example: 2 is a rational number but sqrt(2) is not rational, so 2 is not a perfect square.
It is not possible for a perfect square to have just 2 terms.
Any integer ,n, to the 6th. power would be both a perfect square and a perfect cube: This is because (n2 )3 =n6 which is a perfect square and a perfect cube. Or course you could also write this as (n3 )2 =n6 06 = 0 16 = 1 26 = 64 etc.