If $$\left[x - \left(\frac{1}{x}\right)\right] = 2$$, then what is the value of $$\left[x^6 - \left(\frac{1}{x^6}\right)\right]$$?

$$\left[x-\left(\frac{1}{x}\right)\right]=2$$

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

or, $$x^2+\frac{1}{x^2}-2=4.$$

or, $$x^2+\frac{1}{x^2}=6.$$

or, $$\left(x^2+\frac{1}{x^2}\right)^2=36.$$

or, $$\left(x^4+\frac{1}{x^4}+2\right)=36.$$

or, $$x^4+\frac{1}{x^4}=34..................\left(1\right)$$

Now, 

$$\left(x-\frac{1}{x}\right)=2.$$

or, $$x^2-2x-1=2.$$

or, $$x=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt{2}}{2}=\left(1\pm\sqrt{2}\right).$$

Lets consider positive root value for x.

$$x^2=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}.$$

And, $$\frac{1}{x^2}=\left(\frac{1}{1+\sqrt{2}}\right)^2=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}.$$

So, $$x^2-\frac{1}{x^2}=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}.$$

Now, 

$$\left(x^4+\frac{1}{x^4}\right)\left(x^2-\frac{1}{x^2}\right)=34\times4\sqrt{2}$$

or, $$x^6-\frac{1}{x^6}-\left(x^2-\frac{1}{x^2}\right)=136\sqrt{2}.$$

or, $$x^6-\frac{1}{x^6}-4\sqrt{2}=136\sqrt{2}.$$

or, $$x^6-\frac{1}{x^6}=140\sqrt{2}.$$

D is correct choice.