Question : If $x^2-5 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 21
Option 2: 22
Option 3: 25
Option 4: 24
Correct Answer: 22
Solution : $x^2 - 5x + 1 = 0$ $⇒x(x - 5 + \frac{1}{x}) = 0$ $⇒(x - 5 + \frac{1}{x}) = 0$ $⇒(x + \frac{1}{x}) = 5$ Cubing both sides, we get, $⇒(x + \frac{1}{x})^3 = 5^3$ $\because$ $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ $⇒x^3 + \frac{1}{x^3} + 3.x.\frac{1}{x} (x + \frac{1}{x}) = 125$ $⇒x^3 + \frac{1}{x^3} + 3(5) = 125$ $⇒x^3 + \frac{1}{x^3} + 15 = 125$ $⇒x^3 + \frac{1}{x^3} = 125 - 15=110$ Now, $\frac{x^4 + \frac{1}{x^2}}{x^2 + 1}$ Dividing the numerator and denominator by $x$, $=\frac{\frac{1}{x}(x^4 + \frac{1}{x^2})}{(x^2 + 1)\frac{1}{x}} $ $= \frac{(x^3 + \frac{1}{x^3})}{(x + \frac{1}{x})}$ $=\frac{110}{5}$ $=22$ Hence, the correct answer is 22.
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