Perfect cube factors

Perfect cube factors:
If a number is a perfect cube, then the power of the prime factors should be divisible by 3.
Example 1:Find the number of factors of2⁹3⁶5⁵11⁸ that are perfect cube?
Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have
2^{(0 or 3 or 6or 9)}—– 4 factors
3^{(0 or 3 or 6)}  —–  3  factors
5^{(0 or 3)}——- 2 factors
11^{(0 or 3 or 6 )}— 3 factors
Hence, the total number of factors which are perfect cube 4x3x2x3=72
Perfect square and perfect cube
If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.
Example 2: How many factors of 2⁹3⁶5⁵11⁸ are both perfect square and perfect cube?
Solution: If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have
2^{(0 or 6)}—– 2 factors
3^{(0 or 6)} —– 2 factors
5⁽⁰⁾——- 1 factor
11^{(0 or 6)}— 2 factors
Hence total number of such factors are 2x2x1x2=8
Example 3: How many factors of2⁹3⁶5⁵11⁸are either perfect squares or perfect cubes but not both?
Solution:
Let A denotes set of numbers, which are perfect squares.
If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have
2^{(0 or 2 or 4 or 6 or 8)}—– 5 factors
3^{(0 or 2 or 4 or 6)}  —– 4 factors
5^{(0 or 2or 4 )}——- 3 factors
11^{(0 or 2or 4 or6 or 8 )}— 5 factors
Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300
Let B denotes set of numbers, which are perfect cubes
If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have
2^{(0 or 3 or 6or 9)}—– 4 factors
3^{(0 or 3 or 6)}  —–  3  factors
5^{(0 or 3)}——- 2 factors
11^{(0 or 3 or 6 )}— 3 factors
Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72
If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have
2^{(0 or 6)}—– 2 factors
3^{(0 or 6)} —– 2 factors
5⁽⁰⁾——- 1 factor
11^{(0 or 6)}— 2 factors
Hence total number of such factors are i.e.n(A∩B)=2x2x1x2=8
We are asked to calculate which are either perfect square or perfect cubes i.e.
n(A U B )= n(A) + n(B) – n(A∩B)
=300+72 – 8
=364
Hence required number of factors is 364.