Question : If $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$, and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right) ?$
Option 1: $22970 \sqrt{23}$
Option 2: $23030 \sqrt{23}$
Option 3: $23060 \sqrt{23}$
Option 4: $22960 \sqrt{23}$
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Correct Answer: $23030 \sqrt{23}$
Solution :
Given, $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$
Squaring both sides, we get,
$\left(x+\frac{1}{x}\right)^2=(5 \sqrt{2})^2$
⇒ $x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=25\times 2$
⇒ $x^2+\frac{1}{x^2}+2=50$
⇒ $x^2+\frac{1}{x^2}=50-2$
⇒ $x^2+\frac{1}{x^2}=48$
Subtracting 2 from both sides, we get,
⇒ $(x-\frac{1}{x})^2=48-2$
⇒ $(x-\frac{1}{x})^2=46$
⇒ $(x-\frac{1}{x})=\sqrt{46}$
Now consider, $\left(x^6-\frac{1}{x^6}\right)$
We know, $a^2-b^2=(a-b)(a+b)$
$=(x^3+\frac{1}{x^3})(x^3-\frac{1}{x^3})$
Also, $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$
$=[(x+\frac{1}{x})(x^2-x\times\frac{1}{x}+\frac{1}{x^2})][(x-\frac{1}{x})(x^2+x\times\frac{1}{x}+\frac{1}{x^2})]$
$=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)$
$=(5\sqrt2)(48-1)(\sqrt{46})(48+1)$
$=5\sqrt2\times 47\times \sqrt{46}\times 49$
$=11515\sqrt{2\times 46}$
$=11515\times2\sqrt{23}$
$=23030\sqrt{23}$
Hence, the correct answer is $23030\sqrt{23}$.
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