Trigonometric algebraic expansion

Let $x = \tan A$, then $\cot A = \frac{1}{x}$. Given $x + \frac{1}{x} = \sqrt{5}$. Using the identity $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$: $(\sqrt{5})^3 = x^3 + \frac{1}{x^3} + 3\sqrt{5} \implies 5\sqrt{5} = \tan^3 A + \cot^3 A + 3\sqrt{5}$. $\tan^3 A + \cot^3 A = 2\sqrt{5}$.