Solve $x^3 = 27\pmod {41}$ I don't know how to approach this problem. Can anyone give me a hint? If it matters, the first part of the question was to find the order of $5$ in the field $\mathbb Z_{41}$ (the field mod $41$), which I did, but I'm not sure how it relates to the second part.
Thanks for your help.

We shall use the lemma that if $a^r\equiv a^s\pmod{n}$, then $r\equiv s\pmod{\operatorname{ord}_n(a)}$. Note that $6$ is a primitive root $\pmod{41}$, and that $6^5\equiv 27\pmod{41}$. We can write $x^3 = 6^{3k}$ for some $k$, so that $$6^{3k}\equiv 6^5\pmod{41}\implies 3k\equiv 5\pmod{\varphi(41)=40}.$$ It is straightforward to proceed from here.
