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    If x is the largest and y is the lowest among ∜6,√2 and ∛4, then the value of x3+y2x3−y2\frac{x^3+y^2}{x^3-y^2 }x3−y2x3+y2​​ will be
    Question

    If x is the largest and y is the lowest among ∜6,√2 and ∛4, then the value of x3+y2x3y2\frac{x^3+y^2}{x^3-y^2 }​ will be

    A.

    3

    B.

    3/2

    C.

    2/3

    D.

    1/3

    Correct option is A

    Given:
    Set of numbers = 614,212,413 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:
    614=6312=(63)112=216112 212=2612=(26)112=64112 413=4412=(44)112=2561126^{\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 256112,256^{\frac{1}{12}},​ so x = 413.4^{\frac{1}{3}}.​​
    The lowest is 64112,64^{\frac{1}{12}}, ​so y  =212.= 2^{\frac{1}{2}}.​​
    Substitute x and y into the required expression: x3+y2x3y2.\frac{x^3 + y^2}{x^3 - y^2}.​​
    x3=(413)3=4x^3 = (4^{\frac{1}{3}})^3 = 4​​
    y2=(212)2=2y^2 = (2^{\frac{1}{2}})^2 = 2​​
    Result = 4+242=62=3 \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3​​
    Final Answer
    So the correct answer is (a)

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