Question : If $x^2-4x+1=0$, then the value of $\frac{x^6+1}{x^3}$ is:
Option 1: 48
Option 2: 52
Option 3: 55
Option 4: 58
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Correct Answer: 52
Solution :
Given:
$x^2-4x+1=0$
Dividing both sides by $x$ we get,
⇒ $x-4+\frac{1}{x}=0$
⇒ $x+\frac{1}{x}=4$
⇒ $(x+\frac{1}{x})^3=4^3$
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}×(x+\frac{1}{x})=64$
⇒ $x^3+\frac{1}{x^3}+3×4=64$ [as, $x+\frac{1}{x}=4$]
⇒ $x^3+\frac{1}{x^3}=64-12$
$\therefore \frac{x^6+1}{x^3}=52$
Hence, the correct answer is 52.
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