Given $x^3$ mod $55$, find its inverse So i am wondering how i can figure out what the functional inverse  of $x^3$ mod $55$ is.
I can only assume it is $x^{1/3}$ mod $55$ but i am not sure if that is the form i should keep it in

By the remark below $\,x^{\color{#c00}{\large 20}}\equiv 1\pmod{55}\,$ for all for all $x$ coprime to $55$. Thus all exponents on $x$ can be considered to be $\,$ mod $\,\color{#c00}{20}$.
Solving the equation $\ x^{\large 3}\equiv y\pmod{55}\, $ for $\,x\,$ is easy because $\,1/3\,$ exists mod $20$, namely: 
Raise both sides to power $\,\color{#0a0}{1/3}\equiv 21/3 \equiv\color{#0a0}7\pmod{\!\color{#c00}{20}}$ 
Doing that we obtain $\, x \equiv\, (x^{\large 3})^{\large\color{#0a0}{1/3}}\,\equiv\, y^{\large\color{#0a0}7}\pmod{\color{#c00}{20}} $ 
Remark $\,\color{#c00}{x^4\equiv 1}\,\Rightarrow\,x^{20}\equiv (\color{#c00}{x^4})^5\equiv \color{#c00}{1}^5\equiv 1\pmod{5}\,$ for $x$ coprime to $5$ by little Fermat. Similarly $x^{10}\equiv 1\,\Rightarrow\,x^{20}\equiv 1\pmod{11}\,$ for $x$ coprime to $11$. So for $x$ coprime to $5,11$ we have $x^{20}-1\,$ is divisible by $5$ and $11$ so also by their LCM $= 5\cdot 11,\,$ i.e. $\,x^{20}\equiv 1\pmod{55}\,$  Alternatively, we could use CRT instead of the LCM, but that's a bit overkill.
