Cubic modular equation 
If I found the sol of (a), I could find sol of (b) and (c) by using Hensel's lemma.
I want to know the way of solving (a) without testing all the cases (ex testing 1,2,3,4,...10)
I think that there will be other way not testing all residues. It takes such a long time...
Or is there short cut for testing..?

Note the polynomial  in a) can be written as
$$x^3-3x^2-x-1=(x-1)^3-3x+1-x-1=(x-1)^3-4(x-1)-4.$$
Hence, setting $x-1=y$, it is enough to solve $y^3=4y+4 $.
We'll use Cardano's method: set $y=u+v$. The last equation becomes
$$u^3+v^3+3uv(u+v)=4(u+v)+4,$$
and adding the condition $3uv=4$,  it comes down to
$$\begin{cases}
uv=5\\u^3+v^3=4
\end{cases} \iff \begin{cases}
u^3v^3=4\\u^3+v^3=4
\end{cases} $$
Thus $u^3, v^3$  are solutions to the quadratic equation $\;t^2-4t+4=(t-2)^2=0$: $\;u^3=v^3=2$. If a draw a table of cubes in $\mathbf Z/11\mathbf Z$:
$$\begin{matrix}
x&0&\pm1&\pm2&\pm3&\pm4&\pm5\\\hline
x^3&0&\pm1&\mp3&\pm5&\mp2&\pm4\end{matrix}$$
we see that $u=v=-4$, whence $y=3$ and finally $\color{red}{x=4.}$
To obtain the other roots, divide $y^3-4y-4$ by $y-3$:
$$y^3-4y-4=(y-3)(y^2+3y+5)=(y-3)(y^2-8y+5)=(y-3)(y-4)^2,$$
so the equation in $x$ also has a double root: $\color{red}{x=5.}$
