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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