This is to solve for $m^3$ in an RSA broadcast attack where I have $c1$, $c2$, $c3$, $N1$, $N2$, $N3$ and $e=3$.
I use CRT (Chinese Remainder Theorem) to get $c1 \equiv c2 \equiv c3 \pmod {N_1 N_2 N_3}$ and that is supposedly $m^3$ "unrolled" so $m = \sqrt[3]{m^3}$ and so $m^3$ can be found quickly with the Extended Euclidean Algorithm.
I don't know to piece those two algorithms together and how they work in terms of decrypting $m$ in RSA although I can find the CRT and EEA implementations individually.
I'm presuming you're given $$ c_i\equiv m^3\pmod{N_i}\\ $$ for $\ i=1,2,3\ $, where $\ 0<m,c_i<N_i\ $ for $\ i=1,2,3\ $, and the problem is to recover $\ m\ $. If $\ \gcd(N_i,N_j)=1\ $ for $\ i\ne j\ $ you can use the Chinese remainder construction to find an integer $ \mu\in[0,N_1N_2N_3-1]\ $ (which will be unique) such that $$ \mu\equiv c_i\pmod{N_i} $$ for $\ i=1,2,3\ $. Because $\ m^3\equiv c_i\pmod{N_i}\ $ and $ m^3\in$$\,[0,N_1N_2N_3-1]\ $, then we must have $\ m^3=\mu\ $, and $\ m\ $ must be the unique real cube root of $\ \mu\ $, which you can then calculate efficiently by any number of methods.
There is no separate application of the extended Euclidean algorithm other than its use in the Chinese remainder procedure. The first step in this procedure, for instance, could be to find a positive integer $\ q\ $ such that $$ c_1+qN_1\equiv c_2\pmod{N_2}\ , $$ and the extended Euclidean algorithm is the standard method for doing this.