This is a very trivial question, i seem to have arrived at a proof for an excercise but the proof just doesn't feel.. right. It is too small and simple.
The fact to be proved is that if $AB\equiv AC$ and $A \not\equiv 0$, then $B\equiv C$.
My proof simply uses the identities that $A\equiv A$ always, and if $A \equiv A'$ and $B \equiv B'$, then $AB \equiv A'B'$
Using this fact, i show that $1/a\equiv1/a$ and as already stated, $AB\equiv AC$.
Hence, $(AB)*1/a\equiv(AC) * 1/a$ or $B\equiv C$. Is this correct?
[P.S: I'm sorry, i didn't know how to get the triple equal to sign in here to signify congruence. Editor's note: This now obsolete, JL]
It is simply false modulo $6$.