If t2 - 4t + 1 = 0, then the value of t^3 + 1t^3 is?

Answer: Option C

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