If $x-\frac{1}{x}=3$, find the values of $x^{2}+\frac{1}{x^{2}}$ and $x^{4}+\frac{1}{x^{4}}$.
Let us consider the following equation:
$x-\frac{1}{x}=3$
Squaring both sides, we get:
$\left(x-\frac{1}{x}\right)^{2}=(3)^{2}=9$
$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=9$
$\Rightarrow x^{2}-2 \times x \times \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=9$
$\Rightarrow x^{2}-2+\frac{1}{x^{2}}=9$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=11$ (Adding 2 to both sides)
Squaring both sides again, we get:
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=(11)^{2}=121$
$\Rightarrow\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=121$
$\Rightarrow\left(x^{2}\right)^{2}+2\left(x^{2}\right)\left(\frac{1}{x^{2}}\right)+\left(\frac{1}{x^{2}}\right)^{2}=121$
$\Rightarrow x^{4}+2+\frac{1}{x^{4}}=121$
$\Rightarrow x^{4}+\frac{1}{x^{4}}=119$
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