The product `root3(2).root4(2). root12(32)` equal to

Options

A. `sqrt2`
B. 2
C. `root12(2)`
D. `root12(32)`

Explanation

To solve the problem, we need to find the product of \( \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} \).

### Step-by-Step Solution:

1. **Convert the roots to exponential form**:
\[ \sqrt[3]{2} = 2^{\frac{1}{3}}, \quad \sqrt[4]{2} = 2^{\frac{1}{4}}, \quad \sqrt[12]{32} = 32^{\frac{1}{12}} \]

2. **Express 32 as a power of 2**:
\[ 32 = 2^5 \quad \text{(since } 32 = 2 \times 16 = 2 \times 2^4 = 2^5\text{)} \]
Therefore,
\[ \sqrt[12]{32} = (2^5)^{\frac{1}{12}} = 2^{\frac{5}{12}} \]

3. **Combine all the exponential forms**:
\[ \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot 2^{\frac{5}{12}} \]

4. **Add the exponents**:
Since the bases are the same, we can add the exponents:
\[ 2^{\left(\frac{1}{3} + \frac{1}{4} + \frac{5}{12}\right)} \]

5. **Find a common denominator**:
The least common multiple (LCM) of 3, 4, and 12 is 12. We convert each fraction:
\[ \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12}, \quad \frac{5}{12} = \frac{5}{12} \]

6. **Add the fractions**:
\[ \frac{4}{12} + \frac{3}{12} + \frac{5}{12} = \frac{4 + 3 + 5}{12} = \frac{12}{12} = 1 \]

7. **Final result**:
\[ 2^{1} = 2 \]

Thus, the product \( \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} \) equals \( 2 \).