Answer
See the proof below.
Work Step by Step
Since $ABC=I$, then $|ABC|=|I|=1$
Since $|ABC|=|A||B||C|$ , then $|A||B||C|=1$ and this implies that
$|A|\ne0$ , $ |B|\ne0$ , $|C|\ne0$
Then the matrices $A$, $B$ and $C$ are invertible and then we have that
$(A(BC))^{-1}=I^{-1}$ implies that $(BC)^{-1} A^{-1}=C^{-1}B^{-1} A^{-1}=I $
Thus, by multiplying with $C$ of the last relation, we have that
$ (C C^{-1})(B^{-1} A^{-1})=CI =C $
Thus $ I(B^{-1} A^{-1})=B^{-1} A^{-1}=C $ and then we have, by multiplying with the matrix $A$:
$B^{-1} A^{-1}A=B^{-1} I=B^{-1} =CA$
Therefore, $B^{-1} =CA$
