determining order of $x+1$ given the $x$ has order three I was trying to expand $(x+1)^n$, then plug $x^3$ in to the expansion of the $(x+1)^n$, keep trying it until I get the order, are there any other ways?
So if $x^3\equiv 1\pmod y$, how would I determine order of $x+1$? 
Can I brute force $x$ and $y$ for the condition then just find $x+1\equiv\pmod y$?
$y$ is prime 
 A: Clearly $x\neq1$ and hence
$$
x^3 \equiv 1 \Rightarrow x^2+x+1 \equiv 0 \tag1$$
Now consider
$(x+1)^6$ and show that
$$
(x+1)^6 - 1 \equiv x\,\left(x+2\right)\,\left(x^2+x+1\right)\,\left(x^2+3\,x+3\right)\tag2$$
and conclude that the order is $6$ by showing no smaller power works.
Added in response to comments
Using (1) in (2) we have 
$$(x+1)^6 \equiv 1$$
Hence order of $x+1$ must divide $6$. 


*

*If $x+1$ has order 1 then $x=0$ which is not possible. So the order cannot be 1.

*if $x+1$ has order 2, then from (1) $x+1 \equiv -x^2$, and hence $(x+1)^2 \equiv x^4$ and $x^4 \not \equiv 1$. So the order cannot be 2.

*if $x+1$ has order 3, then from (1) $x+1 \equiv -x^2$, and hence $(x+1)^3 \equiv -x^6 = -1$. This requires $1 \equiv -1$, which means there are only two elements in the multiplicative group but we know that $1$, $x$ and $x^2$ are distinct since order of $x$ is 3. So the order cannot be 3.
Hence the order has to be $6$
A: Key Idea $\,\ x+1 \equiv (-1)x^{-1}$ has order $= {\rm lcm}({\rm ord}(-1),{\rm ord}(x^{-1})) = {\rm lcm}(2,3) = 6$
Suppose the  modulus $ = p\,$ prime. $\ 0 \equiv x^3\!-1 \equiv (x-1)(x^2\!+x+1).\,$ Then $\,x^2\!+x+1\equiv 0\,$ since $\,x\not\equiv 1\,$ (by $\,x\,$ has order $\,3),\ $ thus we can cancel  $\,x-1\not\equiv 0,\,$ being in a field $\ \Bbb Z/p.$
Thus $\ x(x\!+\!1)\equiv -1\ $ so $\, x\!+\!1\equiv -x^{-1}\,$ so $\,x\!+\!1\,$ has order $\,6\,$ by Key Idea $ $ (note $\,-1\,$ has order $\,2,\,$ else it has order $\,1\,$ so $\,-1\equiv 1\,\Rightarrow\, 2\equiv 0,\,$ so $\,x\in \Bbb Z/2\,$ has order $\,3,\,$ contradiction). $\ \ $ QED
Note $\, $ If the modulus  isn't prime then it may fail, e.g. $\,{\rm mod}\ 26\!:\ 3^3\equiv 1\,$ but $\,4^n\not\equiv 1$ for $\,n>0.$
