$x^3=...2017$

one can find $x$ digit by digit, starting from the least significant one. If the least significant one is $a_1$

$a_1^3 \equiv 7 \ (mod 10) \implies a_1=3$

then, every time we find the $j$ -th digit, we solve the equation

$(10^{j}a_{j+1}+\overline{a_j\dots a_1})^3 \equiv 2017 \ mod(10^{j+1})$

for $a_{j+1}$ . Note that you may expand the expression, that is taken to the power 3, in order to solve the equation.

the final result would be $9073$