c
(c) Put the repeated factor

\((x - 1) = y \Rightarrow x = y + 1\)

\(\therefore {{{x^2}} \over {{{(x - 1)}^3}(x - 2)}} = {{{{(1 + y)}^2}} \over {{y^3}(y - 1)}} = {{1 + 2y + {y^2}} \over {{y^3}( - 1 + y)}}\)

Dividing the numerator,

\((1 + 2y + {y^2})\) by \(( - 1 + y)\) till \({y^3}\) appears as factor,

we get \({{1 + 2y + {y^2}} \over { - 1 + y}} = ( - 1 - 3y - 4{y^2}) + {{4{y^3}} \over { - 1 + y}}\)

Given expression = \({{ - 1} \over {{y^3}}} - {3 \over {{y^2}}} - {4 \over y} + {4 \over { - 1 + y}}\)

= \({{ - 1} \over {{{(x - 1)}^3}}} + {{ - 3} \over {{{(x - 1)}^2}}} + {{ - 4} \over {(x - 1)}} + {4 \over {(x - 2)}}\).