Question : If $y+\frac{1}{y}=3$, then what is the value of $\frac{1}{y^3}+y^3 ?$
Option 1: 12
Option 2: 27
Option 3: 24
Option 4: 18
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Correct Answer: 18
Solution : Use: Use the following formula to solve the question: $(y+\frac{1}{y})^3=y^3+\frac{1}{y^3}+3(y+\frac{1}{y})$ ⇒ $y^3+\frac{1}{y^3}=(y+\frac{1}{y})^3-3(y+\frac{1}{y})$ ⇒ $y^3+\frac{1}{y^3}=3^3-3×3$ ⇒ $\frac{1}{y^3}+y^3 = 18$ Hence, the correct answer is 18.
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Question : If $x=p+\frac{1}{p}$ and $y=p-\frac{1}{p}$, then the value of $x^{4}-2x^{2}y^{2}+y^{4}$ is:
Option 1: 24
Option 2: 4
Option 3: 16
Option 4: 8
Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $5\frac{1}{9}$
Option 2: $4\frac{5}{6}$
Option 3: $7\frac{1}{9}$
Option 4: $9\frac{1}{9}$
Question : If $6^{x}=3^{y}=2^{z}$, then what is the value of $\frac{1}{y}+\frac{1}{z}-\frac{1}{x}$?
Option 1: 1
Option 2: 0
Option 3: 3
Option 4: 6
Question : What is the value of $\frac{4x^2+9y^2+12xy}{144}$?
Option 1: $(\frac{x}{3} + \frac{y}{4})^2$
Option 2: $(\frac{x}{3} + y)^2$
Option 3: $(\frac{x}{4} + \frac{y}{6})^2$
Option 4: $(\frac{x}{6} + \frac{y}{4})^2$
Question : If $x^2+\frac{1}{x^2}=7$, then what is the value of $x^3+\frac{1}{x^3}$?
Option 1: 9
Option 2: 18
Option 3: 27
Option 4: 36
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