A divisibility question involving Fermat's theorem use How I am supposed to solve questions like 
if $\gcd (a,42) =1$ and $168= 3\times 7\times8$ then show that $168$ divides $a^6-1$.

By Fermat's theorem, $a^6 \equiv 1 \pmod{7}$ thus $a^6-1 = 7k$ for some $k \in \mathbb{N}$. Likewise, $a^6 = (a^2)^3 \equiv 1 \pmod{3}$ thus $a^6-1 = 3m$. Finally, $a^6-1 = (a-1)(a+1)(a^4+a^2+1)$. One of $a-1$ and $a+1$ is divisible by at least $4$ and the other by at least $2$ for any odd $a$, thus the whole expression is divisible by at least $8$.
