If $${x^2} + x = 5{\text{,}}$$   then the value of $${\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr & {\text{Let, }}t = x + 3,{\text{ }}x = t - 3 \cr & \Rightarrow {x^2} + x = 5{\text{, }}\left( {{\text{Given}}} \right) \cr & \Rightarrow \left( {x + 1} \right)x = 5 \cr & \Rightarrow \left( {t - 3 + 1} \right).\left( {t - 3} \right) = 5 \cr & \Rightarrow \left( {t - 2} \right).\left( {t - 3} \right) = 5 \cr & \Rightarrow {t^2} - 3t - 2t + 6 = 5 \cr & \Rightarrow {t^2} - 5t = - 1 \cr & \Rightarrow t + \frac{1}{t} = 5 \cr & \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} \cr & \Rightarrow \left( {t + \frac{1}{t}} \right)^3 - 3 \times t \times \frac{1}{t}\left( {t + \frac{1}{t}} \right) \cr & \Rightarrow {5^3} - 3 \times 5 \cr & \Rightarrow 125 - 15 \cr & \Rightarrow 110 \cr & \therefore {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} = 110 \cr} $$