Question : If $x^{4}+\frac{1}{x^{4}}=119$, then the values of $x^{3}+\frac{1}{x^{3}}$ are:
Option 1: $\pm 10\sqrt{13}$
Option 2: $\pm \sqrt{13}$
Option 3: $\pm 16\sqrt{13}$
Option 4: $\pm 13\sqrt{13}$
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Correct Answer: $\pm 10\sqrt{13}$
Solution :
Given: $x^4+\frac{1}{x^4}=119$
Adding 2 on both sides,
⇒ $x^4+\frac{1}{x^4}+2=121$
⇒ $(x^2+\frac{1}{x^2})^2=121$
⇒ $(x^2+\frac{1}{x^2})=11,-11$
Since, $(x^2+\frac{1}{x^2})>0$,
⇒ $(x^2+\frac{1}{x^2})=11$
Adding 2 on both sides,
⇒ $x^2+\frac{1}{x^2}+2=13$
⇒ $(x+\frac{1}{x})^2=13$
⇒ $(x+\frac{1}{x})=\pm\sqrt{13}$
Cubing both sides,
⇒ $x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=\pm13\sqrt{13}$
⇒ $x^3+\frac{1}{x^3}+3(\pm\sqrt{13})=\pm13\sqrt{13}$
⇒ $x^3+\frac{1}{x^3}=\pm10\sqrt{13}$
Hence, the correct answer is $\pm10\sqrt{13}$.
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