If 4/3 ( x^2 + 1x^2 ) = 1102/3, find 1/9 ( x^3 - 1x^3 ), where x > 0.

Answer: Option A

$$\begin{array}{rl}& \frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}\\ & x^2+\frac{1}{x^2}=\frac{332}{3}\times \frac{3}{4}\\ & x^2+\frac{1}{x^2}=83\\ & {\left(x+\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}\\ & {\left(x+\frac{1}{x}\right)}^2=83-2\\ & {\left(x+\frac{1}{x}\right)}^2=81\\ & x+\frac{1}{x}=9\\ & \text{Hence,}\\ & \frac{1}{9}\left(x^3-\frac{1}{x^3}\right)\\ & =\frac{1}{9}\left[{\left(x-\frac{1}{x}\right)}^3+3\times \left(x-\frac{1}{x}\right)\right]\\ & =\frac{1}{9}\left[729+3\times 9\right]\\ & =84\end{array}$$