If $${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$$    then find the value of $${a^{30}}$$ + $${a^{24}}$$ + $${a^{18}}$$ + $${a^{12}}$$ + $${a^6}$$ + $$1$$ = ?

Solution (By Examveda Team)

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Cube both sides}} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} + 1 = 0 \cr & \therefore {\text{ }}{a^{30}} + {a^{24}} + {a^{18}} + {a^{12}} + {a^6} + 1 = ? \cr & \Rightarrow {\text{ }}{a^{24}}\left( {{a^6} + 1} \right) + {a^{12}}\left( {{a^6} + 1} \right) + {a^6} + 1 \cr & \Rightarrow {a^{24}}\left( 0 \right) + {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$