Find which of the following are perfect cubes ?
588

Answer

To determine if 588 is a perfect cube, we need to factor it into its prime factors and analyze the powers of these prime factors.

### Step-by-Step Solution:

1. **Factorization of 588**:
- Since 588 is an even number, we can start by dividing it by 2 (the smallest prime number).
- \( 588 \div 2 = 294 \)
- So, we have \( 588 = 2 \times 294 \).

2. **Continue Factoring 294**:
- Again, 294 is even, so we divide by 2.
- \( 294 \div 2 = 147 \)
- Thus, \( 588 = 2^2 \times 147 \).

3. **Factor 147**:
- Now, we need to factor 147. The sum of the digits of 147 is \( 1 + 4 + 7 = 12 \), which is divisible by 3. Therefore, we can divide by 3.
- \( 147 \div 3 = 49 \)
- So, \( 588 = 2^2 \times 3 \times 49 \).

4. **Factor 49**:
- The number 49 is \( 7 \times 7 \) or \( 7^2 \).
- Therefore, we can write \( 588 = 2^2 \times 3^1 \times 7^2 \).

5. **Analyzing the Exponents**:
- The prime factorization of 588 is \( 2^2 \times 3^1 \times 7^2 \).
- For a number to be a perfect cube, the exponents of all prime factors must be multiples of 3.
- Here, the exponents are:
- For 2: \( 2 \) (not a multiple of 3)
- For 3: \( 1 \) (not a multiple of 3)
- For 7: \( 2 \) (not a multiple of 3)

6. **Conclusion**:
- Since none of the exponents (2, 1, and 2) are multiples of 3, we conclude that 588 is **not a perfect cube**.