If $$x^{2}-9x-1=0$$, then what is the value of $$ (x^{2}-\frac{1}{x^{2}}+5x-\frac{5}{x})$$ ?

Given : $$x^{2}-9x-1=0$$

=> $$x^2-1=9x$$

Dividing both sides by $$'x'$$,

=> $$x-\frac{1}{x}=9$$ --------------(i)

Squaring both sides, we get :

=> $$(x-\frac{1}{x})^2=(9)^2$$

=> $$x^2-\frac{1}{x^2}-2(x)(\frac{1}{x})=81$$

=> $$x^2-\frac{1}{x^2}=81+2=83$$ -----------(ii)

$$\therefore$$ $$ (x^{2}-\frac{1}{x^{2}}+5x-\frac{5}{x})$$

= $$ (x^{2}-\frac{1}{x^{2}})+5(x-\frac{1}{x})$$

Substituting values from equation (i) and (ii),

= $$83+5(9)=128$$

=> Ans - (B)