But how can you be so precise?

Assuming $(a \pm b\sqrt 5)^3 = 38 \pm 17\sqrt 5$ , where $a$ and $b$ are positive integers. Then, we have:

$\begin{aligned} (a + b\sqrt 5)^3 & = 38 + 17\sqrt 5 \\ a^3 + 3a^2b\sqrt 5 + 15ab^2 + 5b^3\sqrt 5 & = 38 + 17\sqrt 5 \end{aligned}$

Equating the rational and irrational parts on both sides:

$\begin{cases} a^3 + 15ab^2 = 38 \\ 3a^2b + 5b^3 =17 \end{cases}$

We note that $b$ , a positive integer, can only be 1, that is $b=1$ $\implies a=2$ $\implies \begin{cases} 39+17\sqrt 5= (2+\sqrt 5)^3 \\ 39-17\sqrt 5 = (2-\sqrt 5)^3 \end{cases}$ . Therefore, we have:

$\begin{aligned} \sqrt[3]{39+17\sqrt 5} + \sqrt[3]{39-17\sqrt 5} & = \sqrt[3]{(2+\sqrt 5)^3} + \sqrt[3]{(2-\sqrt 5)^3} \\ & = 2+\sqrt 5 + 2-\sqrt 5 \\ & = 4 \end{aligned}$

Therefore, the 2016th digit after the decimal point is $\boxed{0}$ .

Let $A$ denote the given number. Recalling that $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ and expanding $A^3$ , we get that $A^3 = 76 - 3A$ . The polynomial $x^3 + 3x - 76$ has the root $x=4$ , which is easily seen by using the rational root test. Dividing out by $x-4$ gives $x^2 + 4x + 19$ , and this polynomial has no real roots. Therefore, $x = 4$ is the only real root and $A = 4$ . Clearly all of its digits after the decimal point are zero...