Let a, b, c be nonzero complex numbers such that 

$\mathrm{a}-\frac{1}{\mathrm{b}}=3, \mathrm{b}-\frac{1}{\mathrm{c}}=4 \mathrm{c}-\frac{1}{\mathrm{a}}=5.$

Find $\mathrm{abc}-\frac{1}{\mathrm{abc}}$ .

We introduce some symmetry by multiplying the given relations:

$(\mathrm{a}-\frac{1}{\mathrm{b}})(\mathrm{b}-\frac{1}{\mathrm{c}})(\mathrm{c}-\frac{1}{\mathrm{a}})=60.$

Finally, adding the two previous equalities, we obtain

$\mathrm{abc}-\frac{1}{\mathrm{abc}}=72$