Answer
$$\frac{1}{{5\left( {z - 3} \right)}} - \frac{1}{{5\left( {z + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{z}{{{z^3} - {z^2} - 6z}} \cr
& {\text{Factoring the denominator}} \cr
& = \frac{z}{{z\left( {{z^2} - z - 6} \right)}} \cr
& = \frac{z}{{z\left( {z - 3} \right)\left( {z + 2} \right)}} \cr
& = \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} \cr
& {\text{The partial fraction decomposition has the form}} \cr
& \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{A}{{z - 3}} + \frac{B}{{z + 2}}\,\,\,\left( {\bf{1}} \right) \cr
& {\text{Multiplying by }}z\left( {z - 3} \right)\left( {z + 2} \right){\text{, we have}} \cr
& 1 = A\left( {z + 2} \right) + B\left( {z - 3} \right) \cr
& \cr
& \,\,\,\,\,{\text{set }}z = 3 \cr
& 1 = A\left( {3 + 2} \right) + B\left( {3 - 3} \right) \cr
& A = \frac{1}{5} \cr
& \cr
& \,\,\,\,\,{\text{set }}z = - 2 \cr
& 1 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 3} \right) \cr
& B = - \frac{1}{5} \cr
& \cr
& {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{{1/5}}{{z - 3}} + \frac{{ - 1/5}}{{z + 2}} \cr
& \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{1}{{5\left( {z - 3} \right)}} - \frac{1}{{5\left( {z + 2} \right)}} \cr} $$
