If $$a + \frac{1}{a} + 1 = 0\left( {a \ne 0} \right){\text{,}}$$     then the value of $$\left( {{a^4} - a} \right)\,{\text{is?}}$$

$$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$