So while I was messing around with binomial coefficients I noticed that $$ \binom{3p-1}{p}\equiv 2 \pmod{p^3} $$
For all the primes I tested above 2. I looked around and found similar congruences online, so i'm guessing this isnt that out the ordinary.
However, I also found that $$ \binom{6q-1}{2q}\equiv \sum_{k=0}^{k=2q}\binom{2q}{k}^3\pmod{8q^3} $$ This however, was not true for all primes q, with the exceptions I found being $2$, $5$, $17$, $19$, $37$, $41$, $73$, $137$ and $149$. The remainder $\text{mod } 8q^3$ was not simple in most cases. So does anyone know how one would attempt the second problem, to prove it, or to prove how many exceptions there are, or any pattern among the exceptions. Any help or insight would be much appreciated.