Question : If $x^{4}+\frac{1}{x^{4}}=34$, what is the value of $x^{3}-\frac{1}{x^{3}} $?
Option 1: 0
Option 2: 6
Option 3: 8
Option 4: 14
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Correct Answer: 14
Solution : Given: $x^{4}+\frac{1}{x^{4}}=34$ Adding 2 to both sides, we get, $⇒x^{4}+\frac{1}{x^{4}}+2=34+2$ $⇒(x^{2}+\frac{1}{x^{2}})^{2}=(6)^{2}$ $⇒x^{2}+\frac{1}{x^{2}}=6$ Subtracting 2 from both sides, we get, $⇒x^{2}+\frac{1}{x^{2}}-2=6-2$ $⇒(x-\frac{1}{x})^{2}=(2)^{2}$ $⇒x-\frac{1}{x}=2$ Now, $(x-\frac{1}{x})^3=x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})$ $⇒2^3=x^3-\frac{1}{x^3}-3×2$ $\therefore x^{3}-\frac{1}{x^{3}}=14$ Hence, the correct answer is 14.
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Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Question : If $(x + \frac{1}{x})$ = 6, then the value of ($x^{2} + \frac{1}{x^{2}}$) is:
Question : If $\left(x-\frac{1}{x}\right)^2=12$, what is the value of $\left(x^2-\frac{1}{x^2}\right)$, given that $x>0$?
Question : If $x+\frac{1}{x}=6$, then find the value of $\frac{3 x}{2 x^2-5 x+2}$.
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