Prove that if 3 is the order of x (mod p) then 6 is the order of x+1 (mod p). The hint says factor $\displaystyle x^3 - 1$, which is $\displaystyle (x-1)(x^2+x+1)$, but I don't see any relation between $\displaystyle x^3-1$ and $\displaystyle (x-1)^6$. Some help please.