\(\therefore {(1 + x)^{ - 1}} = {\log _{abc}}a\)
\(\therefore {(1 + x)^{ - 1}} + {(1 + y)^{ - 1}} + {(1 + z)^{ - 1}} = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c\)
\( = {\log _{abc}}abc = 1\).