Which is greater `root3(2)` or `sqrt(3)`?

Options

A. cannot be compared
B. `root3(2)`
C. `sqrt(3)`
D. equal

Answer

To determine which is greater between the cube root of 2 (denoted as \( \sqrt[3]{2} \)) and the square root of 3 (denoted as \( \sqrt{3} \)), we can follow these steps:

### Step 1: Express the roots in terms of exponents
The cube root of 2 can be expressed as:
\[ \sqrt[3]{2} = 2^{1/3} \]
And the square root of 3 can be expressed as:
\[ \sqrt{3} = 3^{1/2} \]

### Step 2: Find a common exponent
To compare \( 2^{1/3} \) and \( 3^{1/2} \), we can convert them to have a common exponent. The least common multiple (LCM) of the denominators (3 and 2) is 6. Therefore, we can rewrite both expressions with an exponent of \( \frac{1}{6} \).

For \( 2^{1/3} \):
\[ 2^{1/3} = 2^{2/6} = (2^2)^{1/6} = 4^{1/6} \]

For \( 3^{1/2} \):
\[ 3^{1/2} = 3^{3/6} = (3^3)^{1/6} = 27^{1/6} \]

### Step 3: Compare the bases
Now we have:
\[ \sqrt[3]{2} = 4^{1/6} \quad \text{and} \quad \sqrt{3} = 27^{1/6} \]
Since both expressions are raised to the same power \( \frac{1}{6} \), we can compare the bases:
- \( 4 \) (from \( 4^{1/6} \))
- \( 27 \) (from \( 27^{1/6} \))

Since \( 27 > 4 \), it follows that:
\[ 27^{1/6} > 4^{1/6} \]

### Conclusion
Thus, we conclude that:
\[ \sqrt{3} > \sqrt[3]{2} \]

### Final Answer
The square root of 3 is greater than the cube root of 2.

---