Question : If $x^2-4 x+1=0$, then what is the value of $\left(x^6+x^{-6}\right)$?
Option 1: 2786
Option 2: 2702
Option 3: 2716
Option 4: 2744
Correct Answer: 2702
Solution : Given: $x^2 - 4x + 1 = 0$ Divided by $x$ ⇒ $x - 4 + \frac{1}{x} = 0$ ⇒ $x + \frac{1}{x} = 4$ As we know, $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$ ⇒ $4^2 = x^2 + \frac{1}{x^2} + 2$ ⇒ $x^2 + \frac{1}{x^2}$ = 16 – 2 = 14 Again as we know, $(x^2 + \frac{1}{x^2})^3 = x^6 + \frac{1}{x^6} + 3(x^2 + \frac{1}{x^2})$ ⇒ $14^3 = x^6 + \frac{1}{x^6} + 3 × 14$ ⇒ $x^6 + x^{-6}$ = 2744 – 42 = 2702 Hence, the correct answer is 2702.
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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
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
Question : If $x(x-5)=-1$, then the value of $x^3\left(x^3-110\right)=$?
Option 1: 0
Option 2: –1
Option 3: 1
Option 4: 2
Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Option 1: 4
Option 2: 5
Option 3: 6
Option 4: 8
Question : If $x^4+y^4=x^2 y^2$, then the value of $x^6+y^6$ is:
Option 1: 2
Option 2: 0
Option 4: 3
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