Why does $x^7 = 5$ (mod 11) simplify into $5x^3 = 1$ (mod 11)?

Question

I was struggling on a question I found in my math book so I looked at the answers which contained:

... We only need to solve $x^7 = 5$ (mod 11) or $5x^3 = 1$ (mod 11) ...

With no further explanation.

Answer 1

\begin{align*} &x^7\equiv 5\;(\text{mod}\;11)\\[4pt] \implies\;&x^3{\,\cdot\,}x^7\equiv 5x^3\;(\text{mod}\;11)\\[4pt] \implies\;&x^{10}\equiv 5x^3\;(\text{mod}\;11)\\[4pt] \therefore\;\;\,&5x^3\equiv 1\;(\text{mod}\;11) &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\text{$ \bigl($by Fermat's little Theorem, which is applicable}\\[0pt] &&&\!\!\!\!\!\!\!\!\!\!\!\!\text{since $x^7\equiv 5\;(\text{mod}\;11)$ implies $x\not\equiv 0\;(\text{mod}\;11)\bigr)$}\\[4pt] \end{align*}