$$cotA = \frac{1}{tanA}$$
let $$tanA = x, then cotA = \frac{1}{x}$$
Given, x + $$\frac{1}{x} = \sqrt 5$$.............................(1)
Cubing on both sides, we get
$$(x + \frac{1}{x})^3 = \sqrt 5^3$$
$$\Rightarrow x^3 + \frac{1}{x}^3 + 3\times x\times \frac{1}{x}\times (x + \frac{1}{x}) = 5\sqrt 5$$
$$\Rightarrow x^3 + \frac{1}{x}^3 + 3\times \sqrt 5 = 5\sqrt 5$$
$$\Rightarrow x^3 + \frac{1}{x}^3 = 2\sqrt 5$$.
$$\Rightarrow tanA^3 + cotA^3 = 2\sqrt 5$$.
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