How to do this step quickly in Chinese remainder theorem I have 
$ \begin{cases} x \equiv 2 \pmod 3 \\ x \equiv 4 \pmod 7 \\ x \equiv 5 \pmod8 \end{cases} $
and I don't know how to do this quickly in this step:
$56x_1 \equiv 1 \pmod 3 $ implies $x_1 = 2$
The question is, how to find $x_1, x_2, x_3$ fast? In case $x_1$:
instead of multiplying 56 by $2, 3, 4, 5 \ldots$ and do like this $56*2=112$ so $112 \div 3 \approx 37$ thus $112-37*3=1$
What if $x_n$ where $n \ge 1$ is not $2$ but some big number? I know there is the extended Euclidean algorithm.

One thing to speed things up would be that it seems like you aren't taking advantage of the fact that you can reduce the number $56$ modulo $3$ without affecting anything.
$$56x\equiv1\bmod 3 \quad\leadsto\quad 2x\equiv 1\bmod 3$$
