Finding solutions to polynomial modular equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$ I'm more or less okay with linear modulo equations, and was wondering how to solve polynomial modulo equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$.
The possible solutions are 2, 7, 9, 11 and 16 (found on an online calculator). What is the general method to solving such equations? A method better than just guess for all $x \leq 19$, since then it would be difficult if the modulo is big. Thanks!

there are six sixth roots of one $\pmod {19},$ those being the three cube roots of $1$ and the three cube roots of $-1.$  For each of these, solve $\frac{x+1}{x} \equiv t \pmod {19}$ where $t^6 \equiv 1 \pmod {19}.$ Or, for $t \neq 1,$ you see
$$ \frac{1}{x} \equiv t - 1 \pmod {19},  $$
$$ x \equiv \frac{1}{t-1} \pmod{19} $$
The cube roots of one are $1,7,11 \pmod {19}.$ For instance, $18 \cdot 19 = 342$ and $7^3 = 343.$ Then $7 \cdot 11 = 77 = 1 + 4 \cdot 19.$
The cube roots of $-1$ are therefore $18, 12, 8 \pmod {19}.$
multiplicative inverses
$$
\begin{array}{cccccccccccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
1 & 10 & 13 & 5 & 4 & 16 & 11 & 12 & 17 & 2 & 8 & 9 & 3 & 15 & 14 & 6 & 9 & 18
\end{array}
$$
