<s>[INST] Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$. 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 $168$?
 [/INST] By Fermat's Theorem, $a^6\equiv 1 \pmod 7$.  Also by Fermat's Theorem, or otherwise, $a^2\equiv 1 \pmod 3$.  Thus $a^6\equiv 1 \pmod 3$.  So far, we have that 
$$a^6\equiv 1 \pmod {21}.$$
But $a$ is odd, so $a^2\equiv 1 \pmod 8$. It follows that 
$$a^6 \equiv 1 \pmod {8}.$$
Now it's over.  
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