Find the cube root 1728.

Answer

To find the cube root of 1728, we can follow these steps:

### Step 1: Factorize 1728
Since 1728 is an even number, we can start dividing it by 2.

- 1728 ÷ 2 = 864
- 864 ÷ 2 = 432
- 432 ÷ 2 = 216
- 216 ÷ 2 = 108
- 108 ÷ 2 = 54
- 54 ÷ 2 = 27

Now, we have 27, which is an odd number. We can factor 27 by dividing it by 3.

- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1

### Step 2: Write the prime factorization
Now we can write the prime factorization of 1728:
- 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
- This can be expressed as: 1728 = 2^6 × 3^3

### Step 3: Apply the cube root
To find the cube root of 1728, we can use the property of exponents:
- Cube root of a number can be expressed as raising that number to the power of \( \frac{1}{3} \).

So, we can write:
- \( \sqrt[3]{1728} = \sqrt[3]{2^6 \times 3^3} \)

### Step 4: Simplify using the properties of exponents
Using the property \( (a^m)^n = a^{m \cdot n} \):
- \( \sqrt[3]{2^6} = 2^{6 \cdot \frac{1}{3}} = 2^2 \)
- \( \sqrt[3]{3^3} = 3^{3 \cdot \frac{1}{3}} = 3^1 \)

### Step 5: Calculate the values
Now we can calculate:
- \( 2^2 = 4 \)
- \( 3^1 = 3 \)

Now multiply these results:
- \( 4 \times 3 = 12 \)

### Conclusion
Thus, the cube root of 1728 is 12.