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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)$}$

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)$}$

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