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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