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

Correct option is C

Given:

​$$d² + \frac{1}{d²}= 18.$$ ​

d > 1/d.    

Formula Used: 

​$$when \ x - \frac{1}{x} = k \\\ \\value \ of \ x^3 - \frac{1}{x} = k^3 + 3k$$​​

Solution:    

$$d² + \frac{1}{d²}= 18.$$​

$$d² + \frac{1}{d²} -2 = 18 - 2$$ 

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

$$d - \frac{1}{d} = 4$$ 

Then the value of 

$$d^3 - \frac{1}{d^3} = 4^3 + 3\times 4$$   

$$d^3 - \frac{1}{d^3} = 64 + 12$$  

$$d^3 - \frac{1}{d^3} = 76$$​