Question : If $x^4+\frac{1}{x^4}=194$ and $x^3+\frac{1}{x^3}=?$
Option 1: 52
Option 2: 58
Option 3: 76
Option 4: 67
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Correct Answer: 52
Solution : $x^4+\frac{1}{x^4}=194$ Adding 2 on both sides, ⇒ $x^4+\frac{1}{x^4}+2=194+2$ ⇒ $(x^2+\frac{1}{x^2})^2=196$ Taking square root on both sides, ⇒ $x^2+\frac{1}{x^2}=14$ Adding 2 on both sides, ⇒ $x^2+\frac{1}{x^2}+2=14+2$ ⇒ $(x+\frac{1}{x})^2=16$ Taking square root on both sides, ⇒ $x+\frac{1}{x}=4$ ⇒ $(x+\frac{1}{x})^3=4^3$ ⇒ $x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=64$ ⇒ $x^3+\frac{1}{x^3}+3×4=64$ ⇒ $x^3+\frac{1}{x^3}=64-12 = 52$ Hence, the correct answer is 52.
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Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{5}$
Question : If $x+\frac{1}{x}=1$, then the value of $\frac{x^2+7x+1}{x^2+11x+1}$:
Option 1: $\frac{3}{4}$
Option 2: $\frac{2}{3}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{1}{4}$
Question : If $x\left(3-\frac{2}{x}\right)=\frac{3}{x}$, then the value of $x^3-\frac{1}{x^3}$ is equal to:
Option 1: $\frac{8}{27}$
Option 2: $\frac{61}{27}$
Option 3: $\frac{62}{27}$
Option 4: $\frac{52}{27}$
Question : The value of $(x^{\frac{1}{3}}+x^{-\frac{1}{3}})(x^{\frac{2}{3}}-1+x^{-\frac{2}{3}})$ is:
Option 1: $x^{-1}+x^{\frac{2}{3}}$
Option 2: $x+x^{-\frac{1}{3}}$
Option 3: $x^{\frac{1}{3}}+x^{–1}$
Option 4: $x+x^{–1}$
Question : If $x^4+y^4+x^2 y^2=17 \frac{1}{16}$ and $x^2-x y+y^2=5 \frac{1}{4}$, then one of the values of $(x-y)$ is:
Option 1: $\frac{5}{2}$
Option 3: $\frac{5}{4}$
Option 4: $\frac{3}{2}$
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