Show that $a^6-1$ is divisible by $168$ whenever $(a,42)=1$.

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.

Hint $\ $ Either apply Carmichael's generalization of Euler-Fermat, or proceed directly via

$$\rm A^{N_j}\equiv 1\ \ (mod\ M_j)\ \Rightarrow\ A^{lcm\ N_j}\equiv 1\ \ (mod\ lcm\ M_j)$$

for $\rm \begin{cases}\rm \:N = (2,2,6)\\ \rm M = (8,3,7)\end{cases}\, $ by CCRT. That's what Andre does. It's worth emphasis it's general form.