Decrypting a message? I would like to ask for a little help about the following problem, i got stuck in it and have no idea how to proceed to get the answer which Wolfram Alpha gives (of course, i am not allowed to use the solution by software).
Person A sends a message to person B s.t $a=b^{3}\in \mathbb{F}_{2038074743}$.
If $a=1933360524$, find $b$. 
I know the question might be quite easy, but i don't understand the approach...
Anybody has an idea? Thank you very much in advance!

Note, that $p:=2038074743$ is prime and $p\equiv 2\pmod{3}$, so $3\mid 2p-1$, hence we can define
$$b=a^{\frac{2p-1}{3}}$$
which using $(a,p)=1$ and Fermat can be shown to fulfill $b^3=a$:
$$b^3=a^{2p-1}=a^{2(p-1)+1}=(a^{p-1})^2\cdot a=a$$
