Question : If $x^2-3 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Option 1: 5
Option 2: 6
Option 3: 9
Option 4: 7
Correct Answer: 6
Solution :
Given: $x^2-3 x+1=0$
⇒ $x^2+1=3x$................................(1)
Dividing both sides by $x$, we get:
⇒ $x+\frac{1}{x}=3$...................(2)
Now, cubing both sides,
⇒ $(x+\frac{1}{x})^3=3^3$
⇒ $x^3+\frac{1}{x^3}+3(x\times\frac{1}{x})(x+\frac{1}{x})=27$
⇒ $x^3+\frac{1}{x^3}+3\times3=27$
⇒ $x^3+\frac{1}{x^3}=18$.....................(3)
$\therefore$ The value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$
$=\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+1\right)}$
$=\frac{\left(x^4+\frac{1}{x^2}\right)}{3x}$
$=\frac{x^4}{3x}+\frac{1}{3x\times x^2}$
$=\frac{x^3}{3}+\frac{1}{3\times x^3}$
$=\frac{1}{3}(x^3+\frac{1}{x^3})$
Putting the values, we get
$=\frac{1}{3}\times18=6$
Hence, the correct answer is 6.
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