Question : If $x^2-9x+1=0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $54$
Option 2: $108$
Option 3: $702$
Option 4: $810$
Correct Answer: $702$
Solution : Given: $x^2-9x+1=0$ Dividing both sides by $x$, we get, $⇒x-9+\frac{1}{x}=0$ $⇒x+\frac{1}{x}=9$ Cubing both sides, we get, $⇒(x+\frac{1}{x})^3=9^3$ $⇒x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=729$ $⇒x^3+\frac{1}{x^3}+3×9=729$ $\therefore x^3+\frac{1}{x^3}=729-27=702$ Hence, the correct answer is $702$.
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Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $\frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$, what is the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$
Option 1: –3
Option 2: 0
Option 3: 15
Option 4: 21
Question : If $x\left(5-\frac{2}{x}\right)=\frac{5}{x}$, then the value of $x^2+\frac{1}{x^2}$ is:
Option 1: $\frac{54}{25}$
Option 2: $\frac{53}{28}$
Option 3: $\frac{53}{27}$
Option 4: $\frac{54}{23}$
Question : If $x^4+x^{-4}=47, x>0$, then what is the value of $x+\frac{1}{x}-2?$
Option 1: 1
Option 3: 5
Option 4: 3
Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
Option 1: $0$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $3$
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