Triple fractions (and more complex fractions)Usually a b c = a c b i.e. b...

The trouble is
$a/(b/c)\ne (a/b)/c$
since $a/(b/c)=\frac{ac}{b}$ and $(a/b)/c=\frac{a}{bc}$. So, without brackets, the notation is ambiguous. Introducing brackets, for example, we could choose between
$a/b/c/d=a(b/c/d{)}^{-1}$
and
$a/b/c/d=(a/b/c)d^{-1}$
The first leads to
$a(b(c/d{)}^{-1}{)}^{-1}=a(b{c}^{-1}d{)}^{-1}=\frac{ac}{bd}$
The second leads to
$(a/b)c^{-1}{d}^{-1}=a{b}^{-1}{c}^{-1}{d}^{-1}=\frac{a}{bcd}$
which is less interesting, it seems to me. The first expression leads to
$a/b/c/d/e=a(b(c/d/e){)}^{-1}=a{\left(\frac{bd}{ce}\right)}^{-1}=\frac{ace}{bd}$
Collecting results, the following interesting pattern emerges:
$a/b=a{b}^{-1}=\frac{a}{b}$
$a/b/c=a{b}^{-1}c=\frac{ac}{b}$
$a/b/c/d=a{b}^{-1}c{d}^{-1}=\frac{ac}{bd}$
$a/b/c/d/e=a{b}^{-1}c{d}^{-1}e=\frac{ace}{bd}$
$a/b/c/d/e/f=a{b}^{-1}c{d}^{-1}e{f}^{-1}=\frac{ace}{bdf}...$
and so on.