If $$\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$$, find $$\frac{1}{9} \left(x^3 - \frac{1}{x^3}\right)$$, where $$x$$ > 0.

Given, $$\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$$

$$\Rightarrow$$  $$\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=\frac{332}{3}$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}=83$$

$$\Rightarrow$$  $$x^2+\frac{1}{x^2}-2=83-2$$

$$\Rightarrow$$  $$\left(x-\frac{1}{x}\right)^2=81$$

$$\Rightarrow$$  $$x-\frac{1}{x}=9$$

$$\Rightarrow$$  $$\left(x-\frac{1}{x}\right)^3=9^3$$

$$\Rightarrow$$  $$x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=729$$

$$\Rightarrow$$  $$x^3-\frac{1}{x^3}-3\left(9\right)=729$$

$$\Rightarrow$$  $$x^3-\frac{1}{x^3}-27=729$$

$$\Rightarrow$$  $$x^3-\frac{1}{x^3}=756$$

$$\Rightarrow$$  $$\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=\frac{1}{9}\left(756\right)$$

$$\Rightarrow$$  $$\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=84$$

Hence, the correct answer is Option C