Cube root of 19683 is

Options

A. 19
B. 23
C. 27
D. 29

Answer

To find the cube root of 19683, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the cube root of the number 19683. This means we are looking for a number \( x \) such that \( x^3 = 19683 \).

2. **Prime Factorization**: We will start by factorizing 19683 into its prime factors. This can be done by dividing the number by prime numbers.

- Divide by 3:
- \( 19683 \div 3 = 6561 \)
- Divide 6561 by 3:
- \( 6561 \div 3 = 2187 \)
- Divide 2187 by 3:
- \( 2187 \div 3 = 729 \)
- Divide 729 by 3:
- \( 729 \div 3 = 243 \)
- Divide 243 by 3:
- \( 243 \div 3 = 81 \)
- Divide 81 by 3:
- \( 81 \div 3 = 27 \)
- Divide 27 by 3:
- \( 27 \div 3 = 9 \)
- Divide 9 by 3:
- \( 9 \div 3 = 3 \)
- Finally, divide 3 by 3:
- \( 3 \div 3 = 1 \)

So, the complete factorization of 19683 is:
\[ 19683 = 3^9 \]

3. **Applying the Cube Root**: Now that we have the prime factorization, we can find the cube root.
\[ \sqrt[3]{19683} = \sqrt[3]{3^9} \]

4. **Using the Property of Exponents**: The cube root can be expressed as:
\[ \sqrt[3]{3^9} = 3^{9/3} = 3^3 \]

5. **Calculating the Result**: Now we calculate \( 3^3 \):
\[ 3^3 = 27 \]

Thus, the cube root of 19683 is **27**.

### Final Answer:
The cube root of 19683 is **27**.

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