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    If d2+1d2=18,d^2+\frac{1}{ d^2} =18,d2+d21​=18,​ and d> 1d\frac{1}{d}d1​​ then the value of d3−1d3{d^3}-\frac{1}{d^3}d3−d31​​ is:&nb
    Question

    If d2+1d2=18,d^2+\frac{1}{ d^2} =18,​ and d> 1d\frac{1}{d}​ then the value of d31d3{d^3}-\frac{1}{d^3}​ is:  

    A.

    91

    B.

    64

    C.

    76

    D.

    81

    Correct option is C

    Given:

    d2+1d2=18.d² + \frac{1}{d²}= 18. ​

    d > 1/d.    

    Formula Used: 

    when x1x=k value of x31x=k3+3kwhen \ x - \frac{1}{x} = k \\\ \\value \ of \ x^3 - \frac{1}{x} = k^3 + 3k​​

    Solution:    

    d2+1d2=18.d² + \frac{1}{d²}= 18.

    d2+1d22=182d² + \frac{1}{d²} -2 = 18 - 2  

    d2+1d22×d×1d=16 (d1d)2=16d² + \frac{1}{d²} - 2\times d \times\frac{1}{d}= 16\\\ \\(d - \frac{1}{d})^2= 16

    d1d=4d - \frac{1}{d} = 4  

    Then the value of 

    d31d3=43+3×4d^3 - \frac{1}{d^3} = 4^3 + 3\times 4    

    d31d3=64+12d^3 - \frac{1}{d^3} = 64 + 12   

    d31d3=76d^3 - \frac{1}{d^3} = 76


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