jmc

The units digit of $17^3-17$ is the same as the units digit as $7^{13}-7$. To find the units digit of $7^{13}$, we look at the first few powers of 7 modulo 10: $$\begin{array}{rl}{7}^0& \equiv 1,\\ 7^1& \equiv 7,\\ 7^2& \equiv 7\cdot 7\equiv 49\equiv 9,\\ 7^3& \equiv 7\cdot 9\equiv 63\equiv 3,\\ 7^4& \equiv 7\cdot 3\equiv 21\equiv 1(\mathrm{mod}10).\end{array}$$

Since $7^4\equiv 1(\mathrm{mod}10)$, the remainders become periodic, with period 4. Since $13\equiv 1(\mathrm{mod}4)$, $7^{13}\equiv 7(\mathrm{mod}10)$, so the units digit of $7^{13}-7$ is $\boxed{0}$.