I have been running into this type of problem a lot:
Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$.
First of all, by Euler's theorem, we have that $a^{\phi(42)}\equiv a^{12}\equiv1\pmod{42}.$ Notice that $a^6a^6\equiv1\pmod{42}\text{ and }168=4\cdot42.$ It follows that $a^{12}\equiv1\pmod{168},$ $a^{12}-a^6\equiv1-a^6\pmod{168},$ $a^6(a^6-1)\equiv1-a^6\pmod{168},$ $a^6-1\equiv a^6(1-a^6)\pmod{168},$ $a^6-1\equiv a^6-a^{12}\pmod{168}.$ I get stuck here: How can I show that the RHS is congruent to zero modulo $168$?