Question : If ${\left(x-\frac{1}{x}\right)=\sqrt{6}}$ and $x > 1$, what is the value of ${\left(x^8-\frac{1}{x^8}\right)}$?
Option 1: $1024 \sqrt{15}$
Option 2: $992 \sqrt{15}$
Option 3: $998 \sqrt{15}$
Option 4: $1012 \sqrt{15}$
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Correct Answer: $992 \sqrt{15}$
Solution :
$(x-\frac{1}{x})=\sqrt{6}$
Squaring both sides,
⇒ $(x^{2} +\frac{1}{x^2}-2)=6$
⇒ $(x^{2} +\frac{1}{x^2})=8$
Squaring both sides,
⇒ $(x^{4}+\frac{1}{x^4}+2)=64$
⇒ $(x^{4}+\frac{1}{x^4}) = 62$
Squaring both sides,
⇒ $(x^{8}+\frac{1}{x^8}+2)=3844$
⇒ $(x^{8}+\frac{1}{x^8})=3842$
⇒ $(x^{8}+\frac{1}{x^8})^2=3842^2$
⇒ $(x^{8}-\frac{1}{x^8})^2=(x^{8}+\frac{1}{x^8})^2 - 4$
⇒ $(x^{8}-\frac{1}{x^8})^2=3842^2 - 2^2$
⇒ $(x^{8}-\frac{1}{x^8})^2=(3842 +2)(3842-2)$
⇒ $(x^{8}-\frac{1}{x^8})^2=3844 \times 3840$
⇒ $(x^{8}-\frac{1}{x^8})^2=4 × 31 × 31 × 4 × 4 × 4 × 4 × 15$
⇒ $(x^{8}-\frac{1}{x^8}) = 31 × 2 × 4 × 4 × \sqrt{15}$
Hence, the correct answer is $992 \sqrt{15}$.
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