If $\left(x-\frac{1}{x}\right)=5$, find the values of
(i) $\left(x^{2}+\frac{1}{x^{2}}\right)$
(ii) $\left(x^{4}+\frac{1}{x^{4}}\right)$.
$(i)\left(x-\frac{1}{x}\right)=5$
$\Rightarrow$ Squaring both the sides:
$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=(5)^{2}$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}-2(x)\left(\frac{1}{x}\right)\right)=25$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)-2=25$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)=25+2$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)=27$
Therefore, the value of $\left(x^{2}+\frac{1}{x^{2}}\right)$ is 27 .
$\left(x^{2}+\frac{1}{x^{2}}\right)=27$
$\Rightarrow$ Squaring both the sides:
$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}-2\left(x^{2}\right)\left(\frac{1}{x^{2}}\right)\right)=(27)^{2}$
$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)-2=729$
$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)=729+2$
$\Rightarrow\left(x^{4}+\frac{1}{x^{4}}\right)=731$
Therefore, the value of $\left(x^{4}+\frac{1}{x^{4}}\right)$ is 731 .