If $\left(x-\frac{1}{x}\right)=4$, find the value of
(i) $\left(x^{2}+\frac{1}{x^{2}}\right)$.
(ii) $\left(x^{4}+\frac{1}{x^{4}}\right)$.
(i) 
Given, $\left(x-\frac{1}{x}\right)=4$
Squaring both the sides:
$\left(x-\frac{1}{x}\right)^{2}=4^{2}$
$\Rightarrow(x)^{2}-2 \times(x) \times\left(\frac{1}{x}\right)+\left(\frac{1}{x}\right)^{2}=16$
$\Rightarrow x^{2}-2+\frac{1}{x^{2}}=16$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=16+2$
$\therefore\left(x^{2}+\frac{1}{x^{2}}\right)=18$
(ii) From the first part:
$\left(x^{2}+\frac{1}{x^{2}}\right)=18$
Squaring both the sides :
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=18^{2}$
$\Rightarrow\left(x^{2}\right)^{2}+2 \times\left(x^{2}\right) \times\left(\frac{1}{x^{2}}\right)+\left(\frac{1}{x^{2}}\right)^{2}=324$
$\Rightarrow x^{4}+2+\frac{1}{x^{4}}=324$
$\Rightarrow x^{4}+\frac{1}{x^{4}}=324-2$
$\therefore x^{4}+\frac{1}{x^{4}}=322$