If t² - 4t + 1 = 0, then the value of $${t^3} + \frac{1}{{{t^3}}}$$  is?

Solution (By Examveda Team)

$$\eqalign{ & {t^2} - 4t + 1 = 0 \cr & \Rightarrow {t^2} + 1 = 4t \cr & \Rightarrow \frac{{{t^2} + 1}}{t} = \frac{{4t}}{t} \cr & \Rightarrow t + \frac{1}{t} = 4 \cr & \left[ {{\text{Take cube both sides}}} \right] \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3.t.\frac{1}{t}\left( {t + \frac{1}{t}} \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3\left( 4 \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 64 - 12 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 52 \cr} $$