Given that= $t^2-4t+1=0$
or $t^2+1=4t$
dividing by $t$ in both side we get;
$t+\frac{1}{t}=4$
taking cube in bothe side we get:
$\implies t^3+\frac{1}{t^3}+3*t*\frac{1}{t}*(t+\frac{1}{t})=64$
$\implies t^3+\frac{1}{t^3}+3*4=64$
$\implies t^3+\frac{1}{t^3}=64-12=52$
Option $C$ is correct here.
$\text{Note :$(a+b)^3=a^3+b^3+3ab(a+b)$}$