If $$x^4 + \frac{1}{x^4}$$​ = 322 and x > 1, then what is the value of $$x^3 - \frac{1}{x^3}$$​  ?

Correct option is C

Given:

​$$x^4 + \frac{1}{x^4} = 322$$​​

Formula Used:

$$(x+y)^2 =x^2+y^2+2xy \\(x-y)^2 =x^2+y^2-2xy \\(x-y)^3 =x^3-y^3-3xy(x-y) \\$$​

Solution:

We now that:

​$$x^4 + \frac{1}{x^4} = 322$$​​

​$$(x^2)^2 + \left(\frac{1}{x^2}\right)^2 + 2 \times (x^2)\left(\frac{1}{x^2}\right) = 322+2$$​  

$$\left(x^2 +\frac{1}{x^2}\right)^2 = (18)^2$$​     

​$$\left(x^2 +\frac{1}{x^2}\right) = 18$$​  

​$$(x)^2 +\left(\frac{1}{x}\right)^2 - 2 \times (x)\left(\frac{1}{x}\right) = 18 -2$$​   

​$$\left(x-\frac{1}{x}\right)^2 = (4)^2$$​  

​$$\left(x-\frac{1}{x}\right) = (4)$$​  

Cubing both sides:

​$$(x)^3 - \left(\frac{1}{x}\right)^3 - 3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right) = (4)^3$$​   

​$$x^3 - \frac{1}{x^3}- 3(4) = 64$$​​

​$$x^3 - \frac{1}{x^3} = 64 +12 = \bf76$$​​