Find the cube root of 1728.

Answer

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

### Step 1: Factor 1728
We start by finding the prime factors of 1728. We can divide by 2 repeatedly until we can no longer divide evenly.

- 1728 ÷ 2 = 864
- 864 ÷ 2 = 432
- 432 ÷ 2 = 216
- 216 ÷ 2 = 108
- 108 ÷ 2 = 54
- 54 ÷ 2 = 27 (27 is not divisible by 2, so we switch to the next prime number, which is 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 = 2^6 × 3^3

### Step 3: Group the factors
To find the cube root, we need to group the factors in sets of three:
- From 2^6, we can take 2^2 (which is 4) out as (2^3)^2.
- From 3^3, we can take 3^1 out as (3^3)^1.

### Step 4: Calculate the cube root
Now we can calculate the cube root:
- Cube root of 1728 = Cube root of (2^6 × 3^3) = (2^2 × 3) = 4 × 3 = 12.

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

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