Question : If $x+\frac{1}{x}=\sqrt{3}$, then the value of $x^{3}+\frac{1}{x^{3}}$ is equal to:
Option 1: $1$
Option 2: $3\sqrt{3}$
Option 3: $0$
Option 4: $3$
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Correct Answer: $0$
Solution : Given: $x+\frac{1}{x}=\sqrt{3}$ Cubing both sides we get $(x+\frac{1}{x})^3=(\sqrt{3})^3$ ⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=(3\sqrt{3})$ ⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}–3(x+\frac{1}{x})$ $\because x+\frac{1}{x}=\sqrt{3}$ Thus, $x^3+\frac{1}{x^3}=3\sqrt{3}–3\sqrt{3} = 0$ Hence, the correct answer is $0$.
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Question : If $x=\sqrt{–\sqrt{3}+\sqrt{3+8 \sqrt{7+4 \sqrt{3}}}}$ where $x > 0$, then the value of $x$ is equal to:
Option 1: 3
Option 2: 4
Option 3: 1
Option 4: 2
Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{11}$
Option 2: $\sqrt{17}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{13}$
Question : If $x=\frac{4\sqrt{ab}}{\sqrt a+ \sqrt b}$, then what is the value of $\frac{x+2\sqrt{a}}{x-2\sqrt a}+\frac{x+2\sqrt{b}}{x-2\sqrt b}$(when $a\neq b$)?
Option 1: 0
Option 2: 2
Option 3: 4
Option 4: $\frac{(\sqrt a+\sqrt b)}{(\sqrt a - \sqrt b)}$
Question : If $x+\frac{1}{x}=\sqrt{3}$, the value of $\left (x^{3}+\frac{1}{x^{3}} \right )$ is:
Option 1: $\sqrt{3}$
Option 2: $\frac{1}{\sqrt{3}}$
Option 4: $1$
Question : If $x^4+\frac{16}{x^4}=15617, x>0$, then find the value of $x+\frac{2}{x}$.
Option 1: $\sqrt{121}$
Option 2: $\sqrt{129}$
Option 3: $\sqrt{123}$
Option 4: $\sqrt{127}$
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