Question : If $x+\frac{1}{x}=2 K$, then what is the value of $x^4+\frac{1}{x^4}$?
Option 1: $16 {K}^4-16 {K}^2-1$
Option 2: $8 {K}^4+4 {K}^2-1$
Option 3: $16 {K}^4-16 {K}^2+2$
Option 4: $16 {K}^4-4 {K}^2-1$
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Correct Answer: $16 {K}^4-16 {K}^2+2$
Solution :
$x+\frac{1}{x}=2 K$
$⇒(x+\frac{1}{x})^2=(2 K)^2$
$⇒x^2+\frac{1}{x^2}+2 = 4K^2$
$⇒x^2+\frac{1}{x^2} = 4K^2 -2$
$⇒(x^2+\frac{1}{x^2})^2 = (4K^2 -2)^2$
$⇒x^4+\frac{1}{x^4}+2 = 16K^4-16K^2+4$
$⇒x^4+\frac{1}{x^4} = 16K^4-16K^2+2$
Hence, the correct answer is $16K^4-16K^2+2$.
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