If x is the largest and y is the lowest among ∜6,√2 and ∛4, then the value of $$\frac{x^3+y^2}{x^3-y^2 }$$​ will be

Correct option is A

Given:
Set of numbers = $$6^{\frac{1}{4}}, 2^{\frac{1}{2}}, 4^{\frac{1}{3}}$$​​
Formula Used:
Convert fractional indices to a common denominator to compare values.
Solution:
The denominators of the exponents are 4, 2, and 3. The LCM is 12.
Rewrite the numbers with 12 as the denominator:
​$$6^{\frac{1}{4}} = 6^{\frac{3}{12}} = (6^3)^{\frac{1}{12}} = 216^{\frac{1}{12}} \\ \ \\2^{\frac{1}{2}} = 2^{\frac{6}{12}} = (2^6)^{\frac{1}{12}} = 64^{\frac{1}{12}}\\ \ \\4^{\frac{1}{3}} = 4^{\frac{4}{12}} = (4^4)^{\frac{1}{12}} = 256^{\frac{1}{12}}$$​​
By comparing the bases, the largest is $$256^{\frac{1}{12}},$$​ so x = $$4^{\frac{1}{3}}.$$​​
The lowest is $$64^{\frac{1}{12}},$$​so y  $$= 2^{\frac{1}{2}}.$$​​
Substitute x and y into the required expression: $$\frac{x^3 + y^2}{x^3 - y^2}.$$​​
​$$x^3 = (4^{\frac{1}{3}})^3 = 4$$​​
​$$y^2 = (2^{\frac{1}{2}})^2 = 2$$​​
Result = $$\frac{4 + 2}{4 - 2} = \frac{6}{2} = 3$$​​
Final Answer
So the correct answer is (a)