Question : If $x>0$ and $x^4+\frac{1}{x^4}=254$, what is the value of $x^5+\frac{1}{x^5}?$
Option 1: $717 \sqrt{2}$
Option 2: $723 \sqrt{2}$
Option 3: $720 \sqrt{2}$
Option 4: $726 \sqrt{2}$
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Correct Answer: $717 \sqrt{2}$
Solution : Given: $x^4+\frac{1}{x^4}=254$ ⇒ $x^4+\frac{1}{x^4}+2=254+2$ ⇒ $(x^2+\frac{1}{x^2})^2=256$ ⇒ $x^2+\frac{1}{x^2}=16$ Adding 2 on both sides, ⇒ $x^2+\frac{1}{x^2}+2=18$ ⇒ $(x+\frac{1}{x})^2=18$ ⇒ $x+\frac{1}{x}=\sqrt{18}=3\sqrt{2}$ Cubing on both sides ⇒ $x^3+\frac{1}{x^3}+3(3\sqrt{2})=(3\sqrt{2})^3$ ⇒ $x^3+\frac{1}{x^3}=54\sqrt{2}–9\sqrt{2}$ ⇒ $x^3+\frac{1}{x^3}=45\sqrt{2}$ So, $x^5+\frac{1}{x^5}=(x^3+\frac{1}{x^3})(x^2+\frac{1}{x^2})–(x+\frac{1}{x})$ ⇒ $x^5+\frac{1}{x^5}=(45\sqrt{2})(16)–(3\sqrt{2})$ ⇒ $x^5+\frac{1}{x^5}=720\sqrt{2}–3\sqrt{2}$ ⇒ $x^5+\frac{1}{x^5}=717\sqrt{2}$ Hence, the correct answer is $717\sqrt{2}$.
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