Question : If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to:
Option 1: 18
Option 2: 12
Option 3: 15
Option 4: 16
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Correct Answer: 18
Solution : If $x^2+\frac{1}{x^2}=7$, then the value of $x^3+\frac{1}{x^3}$ where x > 0 is equal to: $(x+\frac{1}{x})^2 = x^2+2+ (\frac{1}{x})^2$ ⇒ $(x+\frac{1}{x})^2 = 2+7$ ⇒ $(x+\frac{1}{x})^2 = 9$ ⇒ $(x+\frac{1}{x}) = 3$ ⇒ $(x+\frac{1}{x})^3 = 3^3$ ⇒ $x^3 + (\frac{1}{x})^3+3x\frac{1}{x}(x+\frac{1}{x}) = 27$ ⇒ $x^3 + (\frac{1}{x})^3+3(3) = 27$ ⇒ $x^3 + (\frac{1}{x})^3 = 27-9$ ⇒ $x^3 + (\frac{1}{x})^3 = 18$ Hence, the correct answer is 18.
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Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
Option 1: $- \frac{7}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{3}{7}$
Option 4: $-\frac{3}{7}$
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
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 $2 x^2-7 x+5=0$, then what is the value of $x^2+\frac{25}{4 x^2} ?$
Option 1: $5 \frac{1}{2}$
Option 2: $7 \frac{1}{4}$
Option 3: $9 \frac{1}{2}$
Option 4: $9 \frac{3}{4}$
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