Question : What is the value of $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$?
Option 1: $\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}$
Option 2: $\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}$
Option 3: $\mathrm{k}^8-\frac{1}{\mathrm{k}^8}$
Option 4: $\mathrm{k}^8+\frac{1}{\mathrm{k}^8}$
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Correct Answer: $\mathrm{k}^8-\frac{1}{\mathrm{k}^8}$
Solution : $(a+b)(a-b) =a^2-b^2$ $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$ $= (k^2-\frac{1}{k^2})\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$ $=(k^4-\frac{1}{k^4})\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$ $= (k^8-\frac{1}{k^8})$ Hence, the correct answer is $(k^8-\frac{1}{k^8})$.
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Question : If $x+\frac{1}{x}=\mathrm{\frac{K}{2}}$, then what is the value of $\frac{x^8+1}{x^4} ?$
Question : What is the value of $\left(k-\frac{1}{k}\right)\left(k^2+\frac{1}{k^2}\right)\left(k^4+\frac{1}{k^4}\right)\left(k^8+\frac{1}{k^8}\right)\left(k^{16}+\frac{1}{k^{16}}\right) ?$
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Question : If $K+\frac{1}{K}=-3$, then what is the value of $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$?
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