If $$a - \frac{1}{{a - 5}} = 10$$   , then the value of $${\left( {a - 5} \right)^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}$$    is:

$$a - \frac{1}{{a - 5}} = 10$$
We can write it is as (subtracting 5 from both sides)
$$\eqalign{ & a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr & {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$
So, take the cube of this equation,
$$\eqalign{ & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$