Which of the following number is a prefect cube?

Options

A. 1525
B. 1728
C. 1458
D. 3993

Answer

To determine which of the given numbers is a perfect cube, we will perform the prime factorization of each number and check if all prime factors can be grouped into triplets.

### Step-by-Step Solution:

1. **Check the first number: 1525**
- Perform prime factorization:
- 1525 ÷ 5 = 305
- 305 ÷ 5 = 61
- 61 is a prime number.
- Prime factorization of 1525: \( 5^2 \times 61^1 \)
- Since there are no triplets formed (the exponents are not multiples of 3), **1525 is not a perfect cube**.

2. **Check the second number: 1728**
- Perform prime factorization:
- 1728 ÷ 2 = 864
- 864 ÷ 2 = 432
- 432 ÷ 2 = 216
- 216 ÷ 2 = 108
- 108 ÷ 2 = 54
- 54 ÷ 2 = 27
- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1
- Prime factorization of 1728: \( 2^6 \times 3^3 \)
- Rewriting \( 2^6 \) as \( (2^2)^3 \) gives us \( 2^2 \) triplet and \( 3^3 \) is already a triplet.
- Since all prime factors can be grouped into triplets, **1728 is a perfect cube**.

3. **Check the third number: 1458**
- Perform prime factorization:
- 1458 ÷ 2 = 729
- 729 ÷ 3 = 243
- 243 ÷ 3 = 81
- 81 ÷ 3 = 27
- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1
- Prime factorization of 1458: \( 2^1 \times 3^6 \)
- Here, \( 2^1 \) does not form a triplet, while \( 3^6 \) can be grouped into two triplets.
- Since not all prime factors can be grouped into triplets, **1458 is not a perfect cube**.

4. **Check the fourth number: 3993**
- Perform prime factorization:
- 3993 ÷ 3 = 1331
- 1331 ÷ 11 = 121
- 121 ÷ 11 = 11
- 11 ÷ 11 = 1
- Prime factorization of 3993: \( 3^1 \times 11^3 \)
- Here, \( 3^1 \) does not form a triplet, while \( 11^3 \) is a triplet.
- Since not all prime factors can be grouped into triplets, **3993 is not a perfect cube**.

### Conclusion:
The only number that is a perfect cube among the options provided is **1728**.