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Solution
We haveĀ $$x^4+\frac{1}{x^4}\ =\ 119$$
we getĀ $$x^4+\frac{1}{x^4}+2=\ 121$$
so we getĀ $$\left(x^2+\frac{1}{x^2}\right)^{^2}=\ 121$$
we getĀ $$x^2+\frac{1}{x^2}=\ 11$$
so we getĀ $$x^2+\frac{1}{x^2}-2\ =\ 9$$
so we get $$\left(x-\frac{1}{x}\right)^{^2}=\ 9$$
so we get $$x-\frac{1}{x}=\ 3$$
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