Find which of the following are perfect cubes ?
1728

Answer

To determine whether 1728 is a perfect cube, we will follow these steps:

### Step 1: Prime Factorization
We start by performing the prime factorization of 1728.

1. Since 1728 is an even number, we can divide it by 2:
- \( 1728 \div 2 = 864 \)
2. We continue dividing by 2:
- \( 864 \div 2 = 432 \)
- \( 432 \div 2 = 216 \)
- \( 216 \div 2 = 108 \)
- \( 108 \div 2 = 54 \)
- \( 54 \div 2 = 27 \)

Now, we have:
\[ 1728 = 2^6 \times 27 \]

3. Next, we factor 27:
- \( 27 = 3 \times 3 \times 3 = 3^3 \)

So, we can rewrite 1728 as:
\[ 1728 = 2^6 \times 3^3 \]

### Step 2: Check for Perfect Cube
To check if 1728 is a perfect cube, we need to see if all the exponents in its prime factorization are multiples of 3.

1. The exponent of 2 is 6, which is a multiple of 3 (since \( 6 \div 3 = 2 \)).
2. The exponent of 3 is 3, which is also a multiple of 3 (since \( 3 \div 3 = 1 \)).

Since both exponents are multiples of 3, we conclude that 1728 is a perfect cube.

### Step 3: Find the Cube Root
To find the cube root of 1728, we can use the prime factorization:

\[ \text{Cube root of } 1728 = \sqrt[3]{2^6 \times 3^3} \]

Using the property of exponents:
\[ \sqrt[3]{a^m} = a^{m/3} \]

We can calculate:
\[ \sqrt[3]{2^6} = 2^{6/3} = 2^2 = 4 \]
\[ \sqrt[3]{3^3} = 3^{3/3} = 3^1 = 3 \]

Thus, the cube root of 1728 is:
\[ 4 \times 3 = 12 \]

### Conclusion
Since we found that \( 1728 = 12^3 \), we conclude that 1728 is indeed a perfect cube.

### Final Answer
1728 is a perfect cube, and its cube root is 12.

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