If $$x^{4}+(\frac{1}{x^{4}})=34$$, then what is the value of $$x^{3}-(\frac{1}{x^{3}})$$ ?

Given : $$x^4+\frac{1}{x^4}=34$$

Adding 2 on both sides,

=> $$x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=34+2$$

=> $$(x^2+\frac{1}{x^2})^2=36$$

=> $$(x^2+\frac{1}{x^2})=\sqrt{36}=6$$

Now subtracting 2 from both sides, we get :

=> $$(x)^2+(\frac{1}{x})^2-2(x)(\frac{1}{x})=6-2$$

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

Cubing both sides, we get :

=> $$(x-\frac{1}{x})^3=(2)^3$$

=> $$x^3-\frac{1}{x^3}-3(x)(\frac{1}{x})(x-\frac{1}{x})=8$$

=> $$x^3-\frac{1}{x^3}-3(2)=8$$

=> $$x^3-\frac{1}{x^3}=8+6=14$$

=> Ans - (D)