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

84

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

find $\frac{1}{9}(x^3-\frac{1}{x^3})$

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

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

= x - \(\frac{1}{x}\) = \(\sqrt {83-2}\) = 9

Now,

x³ - \(\frac{1}{x^3}\) = 9³ + 3 × 9 = 756

$\frac{1}{9}(x^3-\frac{1}{x^3})$ = $\frac{1}{9}(756)$ = 84