Find the cube-roots of :
1728

Answer

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

### Step 1: Factor the number 1728
We begin by finding the prime factorization of 1728.

- Divide by 2:
- 1728 ÷ 2 = 864
- 864 ÷ 2 = 432
- 432 ÷ 2 = 216
- 216 ÷ 2 = 108
- 108 ÷ 2 = 54
- 54 ÷ 2 = 27
- Now, 27 is not divisible by 2, so we divide by 3:
- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1

Thus, the prime factorization of 1728 is:
\[ 1728 = 2^6 \times 3^3 \]

### Step 2: Write the cube root expression
We need to find the cube root of 1728:
\[ \sqrt[3]{1728} = \sqrt[3]{2^6 \times 3^3} \]

### Step 3: Apply the cube root to the factors
Using the property of exponents, we can separate the cube root:
\[ \sqrt[3]{1728} = \sqrt[3]{2^6} \times \sqrt[3]{3^3} \]

### Step 4: Simplify the cube roots
Now we simplify each part:
- For \( \sqrt[3]{2^6} \):
\[ \sqrt[3]{2^6} = 2^{6/3} = 2^2 = 4 \]

- For \( \sqrt[3]{3^3} \):
\[ \sqrt[3]{3^3} = 3^{3/3} = 3^1 = 3 \]

### Step 5: Multiply the results
Now we multiply the results from the previous step:
\[ \sqrt[3]{1728} = 4 \times 3 = 12 \]

### Final Answer
Thus, the cube root of 1728 is:
\[ \sqrt[3]{1728} = 12 \]

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