Question : If $x+\frac{1}{x}=2$, then $x^3+\frac{1}{x^3}=?$
Option 1: 1
Option 2: 8
Option 3: 2
Option 4: 0
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Correct Answer: 2
Solution : $x+\frac{1}{x}=2$ Cubing both sides, we get, $(x+\frac{1}{x})^3=2^3$ ⇒ $x^3 + \frac{1}{x^3}+3\times x\times\frac{1}{x}(x+\frac{1}{x})=8$ ⇒ $x^3 + \frac{1}{x^3} + 3(x+\frac1x)=8$ ⇒ $x^3 + \frac{1}{x^3} +3\times 2=8$ ⇒ $x^3+\frac{1}{x^3}=8-6$ $\therefore x^3+\frac{1}{x^3}=2$ Hence, the correct answer is 2.
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Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
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Question : If $(x+\frac{1}{x})\neq 0$ and $(x^3+\frac{1}{x^3})= 0$, then the value $(x+\frac{1}{x})^4$ is:
Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
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