find the values of

(i) $\left(x+\frac{1}{x}\right)=4$

Squaring both the sides:

$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=(4)^{2}$

$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}+2(x)\left(\frac{1}{x}\right)\right)=16$

$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)+2=16$

$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)=16-2$

$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)=14$

Therefore, the value of $x^{2}+\frac{1}{x^{2}}$ is 14 .

$\left(x^{2}+\frac{1}{x^{2}}\right)=14$

Squaring both the sides:

$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}+2\left(x^{2}\right)\left(\frac{1}{x^{2}}\right)\right)=(14)^{2}$

$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)+2=196$

$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)=196-2$

$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)=194$

Therefore, the value of $x^{4}+\frac{1}{x^{4}}$ is 194 .