Question : If $x=\sqrt2+1$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Option 1: $8\sqrt2$
Option 2: $18\sqrt2$
Option 3: $6\sqrt2$
Option 4: $24\sqrt2$
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Correct Answer: $24\sqrt2$
Solution : Given: $x=\sqrt{2}+1$ Thus, $\frac{1}{x}=\sqrt{2}-1$ $x+\frac{1}{x}=(\sqrt{2}+1)+(\sqrt{2}-1)=(\sqrt{2}+1+\sqrt{2}-1)=2\sqrt{2}$ $x-\frac{1}{x}=(\sqrt{2}+1)-(\sqrt{2}-1)=(\sqrt{2}+1-\sqrt{2}+1)=2$ So, $(x^2+\frac{1}{x^2})=6$ We know that, $x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=(x^2+\frac{1}{x^2})(x+\frac{1}{x})(x-\frac{1}{x})$ Putting the values we get, ⇒ $x^4-\frac{1}{x^4}=6×2×2\sqrt{2}$ $\therefore x^4-\frac{1}{x^4}=24\sqrt{2}$ Hence, the correct answer is $24\sqrt{2}$.
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Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Option 1: $\frac{3}{8}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{4}{9}$
Option 4: $\frac{11}{13}$
Question : If $x+\frac{16}{x}=8$, then the value of $x^2+\frac{32}{x^2}$ is:
Option 1: 20
Option 2: 24
Option 3: 16
Option 4: 18
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Option 1: $\frac{13}{4}$
Option 2: $\frac{13}{8}$
Option 3: $\frac {17}{4}$
Option 4: $\frac{17}{8}$
Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{5}{8}$
Option 4: $\frac{7}{8}$
Question : When $2x+\frac{2}{x}=3$, then the value of ($x^3+\frac{1}{x^3}+2)$ is:
Option 1: $\frac{2}{7}$
Option 2: $\frac{7}{8}$
Option 3: $\frac{7}{2}$
Option 4: $\frac{8}{7}$
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