Compare :
(i) `root(6 )(15) and root(4)(12)" (ii) "sqrt(24) and root(3)(25)`

Answer

To compare the given pairs of numbers, we will follow a systematic approach by converting the roots into powers and then comparing them.

### Problem (i): Compare \( \sqrt[6]{15} \) and \( \sqrt[4]{12} \)

1. **Convert to Exponential Form**:
- Let \( a = \sqrt[6]{15} = 15^{1/6} \)
- Let \( b = \sqrt[4]{12} = 12^{1/4} \)

2. **Equalize the Exponents**:
- To compare \( a \) and \( b \), we can raise both to the power of 12 (the least common multiple of 6 and 4):
\[ a^{12} = (15^{1/6})^{12} = 15^2 = 225 \]
\[ b^{12} = (12^{1/4})^{12} = 12^3 = 1728 \]

3. **Compare the Results**:
- Now we compare \( 225 \) and \( 1728 \):
\[ 225 < 1728 \]
Therefore, \( a < b \).

4. **Conclusion**:
\[ \sqrt[6]{15} < \sqrt[4]{12} \]

### Problem (ii): Compare \( \sqrt{24} \) and \( \sqrt[3]{25} \)

1. **Convert to Exponential Form**:
- Let \( c = \sqrt{24} = 24^{1/2} \)
- Let \( d = \sqrt[3]{25} = 25^{1/3} \)

2. **Equalize the Exponents**:
- To compare \( c \) and \( d \), we can raise both to the power of 6 (the least common multiple of 2 and 3):
\[ c^{6} = (24^{1/2})^{6} = 24^3 \]
\[ d^{6} = (25^{1/3})^{6} = 25^2 = 625 \]

3. **Calculate \( 24^3 \)**:
- Now we calculate \( 24^3 \):
\[ 24^3 = 24 \times 24 \times 24 = 13824 \]

4. **Compare the Results**:
- Now we compare \( 13824 \) and \( 625 \):
\[ 13824 > 625 \]
Therefore, \( c > d \).

5. **Conclusion**:
\[ \sqrt{24} > \sqrt[3]{25} \]

### Final Answers:
1. \( \sqrt[6]{15} < \sqrt[4]{12} \)
2. \( \sqrt{24} > \sqrt[3]{25} \)