If $$\left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154,}}$$     then the value of $$\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$$   is = ?

Solution(By Examveda Team)

$$\eqalign{ & \left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154}} \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = 1156 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = 36 \cr & \Rightarrow a + \frac{1}{a} = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^3} = {6^3} = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 6 = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 216 - 18 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 198 \cr} $$