Question : If $x^2-8 x-1=0$, what is the value of $x^2+\frac{1}{x^2}?$
Option 1: 68
Option 2: 62
Option 3: 64
Option 4: 66
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Correct Answer: 66
Solution : $x^2-8 x-1=0$ ⇒ $x^2 -1 = 8x$ Dividing both sides by $x$ $x-\frac{1}{x} = 8$ Squaring both sides, $(x-\frac{1}{x})^2 = 8^2$ ⇒ $x^2 + \frac{1}{x^2} - 2×x×\frac{1}{x} = 64$ $\therefore$ $x^2 + \frac{1}{x^2} = 64+2=66$ Hence, the correct answer is 66.
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Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Option 1: $\frac{3}{8}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{4}{9}$
Option 4: $\frac{11}{13}$
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Option 1: $\frac{13}{4}$
Option 2: $\frac{13}{8}$
Option 3: $\frac {17}{4}$
Option 4: $\frac{17}{8}$
Question : If $x^{4}+\frac{1}{x^{4}}=34$, what is the value of $x^{3}-\frac{1}{x^{3}} $?
Option 1: 0
Option 2: 6
Option 3: 8
Option 4: 14
Question : If $x^2+\frac{1}{x^2}=66$, the value of $x-\frac{1}{x}$ is:
Option 1: 10
Option 2: 8
Option 3: 9
Option 4: 6
Question : If $x^{2} -3x +1=0$, then the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{\left(x^2+5 x+1\right)}$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{27}{8}$
Option 3: $\frac{5}{2}$
Option 4: $2$
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