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: $\mathrm{k}^8+\frac{1}{\mathrm{k}^8}$

Question : If $K+\frac{1}{K}=-3$, then what is the value of $\left(\frac{K^6+1}{K^3}\right)+\left(\frac{K^4+1}{K^2}\right)$?

Option 1: 27

Option 2: – 29

Option 3: 29

Option 4: – 27

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: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2+32}{16}$

Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.

Option 1: $5\frac{1}{9}$

Option 2: $4\frac{5}{6}$

Option 3: $7\frac{1}{9}$

Option 4: $9\frac{1}{9}$

Question : What is the value of $\frac{1+\mathrm{x}}{1-\mathrm{x}^2} \div \frac{1+\mathrm{x}}{1-\mathrm{x}^4}-\frac{1-\mathrm{x}^4}{1-\mathrm{x}} \times \frac{1+\mathrm{x}}{1-\mathrm{x}^2}$?

Option 1: $\frac{2 \mathrm{x}\left(1+\mathrm{x}^2\right)}{(1-\mathrm{x})}$

Option 2: $\frac{2 \mathrm{x}\left(1+\mathrm{x}^2\right)}{(\mathrm{x}-1)}$

Option 3: $(1-\mathrm{x})^2$

Option 4: $\left(1+\mathrm{x}^2\right)$