Question : If $x+\frac{1}{x}=\mathrm{\frac{K}{2}}$, then what is the value of $\frac{x^8+1}{x^4} ?$
Option 1: $\frac{\mathrm{K}^4-16 \mathrm{~K}^2+32}{16}$
Option 2: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2-36}{32}$
Option 3: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2-32}{16}$
Option 4: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2+32}{16}$
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Correct Answer: $\frac{\mathrm{K}^4-16 \mathrm{~K}^2+32}{16}$
Solution :
$x + \frac{1}{x} = \frac{K}{2}$
$⇒\left(x + \frac{1}{x}\right)^2 = \mathrm{\left(\frac{K}{2}\right)^2}$
$⇒x^2 + 2 + \frac{1}{x^2} = \mathrm{\frac{K^2}{4}}$
$⇒x^2 + \frac{1}{x^2} =\mathrm{ \frac{K^2}{4} - 2}$
Now, square this equation:
$⇒\left(x^2 + \frac{1}{x^2}\right)^2 = \mathrm{\left(\frac{K^2}{4} - 2\right)^2}$
$⇒x^4 + 2 + \frac{1}{x^4} = \mathrm{\left(\frac{K^2}{4} - 2\right)^2}$
$⇒x^4 + \frac{1}{x^4} = \mathrm{\left(\frac{K^2}{4} - 2\right)^2 - 2}$
$⇒\frac{x^8+1}{x^4} =\mathrm{(\frac{K^4}{16}-K^2+4)-2}$
$⇒\frac{x^8+1}{x^4} =\mathrm{(\frac{K^4}{16}-K^2+2)}$
$\therefore \frac{x^8+1}{x^4} =\mathrm{(\frac{K^4-16K^2+32}{16})}$
Hence, the correct answer is $\mathrm{(\frac{K^4-16K^2+32}{16})}$.
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