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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Question : The value of $\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}+\left(2 \div 2\right.$ of $\left.\frac{1}{2}\right)$ is:
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