Question : If $t^{2}-4t+1=0,$ then the value of $t^3+\frac{1}{t^3}$ is:
Option 1: 44
Option 2: 48
Option 3: 52
Option 4: 64
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Correct Answer: 52
Solution :
Given: $t^2 - 4t + 1 = 0$
⇒ $t^2 + 1 = 4t$
⇒ $\frac{t^2 + 1}{t} = 4$
⇒ $t + \frac{1}{t} = 4$
On cubing both sides,
$(\frac{t + 1}{t})^3 = 4^3$
Using identity: $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$,
We get: $t^3 + \frac{1}{t^3} + 3(t+\frac{1}{t})= 4^3$
⇒ $ t^3 + \frac{1}{t^3}+3(4)=64$
⇒ $t^3 + \frac{1}{t^3}= 64-12=52$
Hence, the correct answer is 52.
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