If $${\left( {a + \frac{1}{a}} \right)^2}\, = 3$$    then $${a^3} + \frac{1}{{{a^3}}} = ?$$

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & a + \frac{1}{a} = \sqrt 3 \cr & {\text{Take cube on both sides}} \cr & {\left( {a + \frac{1}{a}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$