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: $\frac{17}{8}$
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Correct Answer: $\frac{13}{8}$
Solution : Given: $2x-\frac{2}{x}=1$ Dividing both sides by 2, we get, ⇒ $x-\frac{1}{x} = \frac{1}{2}$ Cubing both sides, we get, ⇒ $(x-\frac{1}{x})^3=(\frac{1}{2})^3$ ⇒ $x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})=\frac{1}{8}$ ⇒ $x^3-\frac{1}{x^3}-3×\frac{1}{2}=\frac{1}{8}$ ⇒ $x^3-\frac{1}{x^3}=\frac{1}{8}+\frac{3}{2}$ $\therefore x^3-\frac{1}{x^3}=\frac{13}{8}$ Hence, the correct answer is $\frac{13}{8}$.
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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 $2x^2+5 x+1=0$, then one of the values of $x-\frac{1}{2 x}$ is:
Option 1: $\frac{\sqrt{17}}{2}$
Option 2: $\frac{13}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{\sqrt{13}}{2}$
Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{5}{8}$
Option 4: $\frac{7}{8}$
Question : When $2x+\frac{2}{x}=3$, then the value of ($x^3+\frac{1}{x^3}+2)$ is:
Option 1: $\frac{2}{7}$
Option 2: $\frac{7}{8}$
Option 3: $\frac{7}{2}$
Option 4: $\frac{8}{7}$
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:
Option 1: 9
Option 2: 12
Option 3: 15
Option 4: 16
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