If $$a - \frac{1}{{a - 3}} = 5{\text{,}}$$   then the value of $${\left( {a - 3} \right)^3}$$   - $$\frac{1}{{{{\left( {a - 3} \right)}^3}}} = ?$$

$$\eqalign{ & a - \frac{1}{{a - 3}} = 5 \cr & \Rightarrow a - 3 - \frac{1}{{a - 3}} = 5 - 3 \cr & \Rightarrow \left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}} = 2 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]^3} = {\left( 2 \right)^3} \cr} $$
  $$ \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3 \times {\left( {a - 3} \right)} \times $$       $$\frac{1}{{{{\left( {a - 3} \right)}}}}$$ $$\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]$$    $$ = 8$$
$$\eqalign{ & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3\left( 2 \right) = 8 \cr & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 8 + 6 \cr & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 14 \cr} $$