Solution to two non polynomial equation If
$$x+\frac1x=-1$$
Find the value of
$$x^{99}+\frac{1}{x^{99}}$$
Is there any formal (traditional) way to solve these problems using only elementary algebra or high school math

$$(x+\frac{1}{x})(x^2+\frac{1}{x^2})\Rightarrow x^3+\frac{1}{x^3}=2$$
$$(x+\frac{1}{x})(x^3+\frac{1}{x^3})\Rightarrow x^4+\frac{1}{x^4}=-1$$
$$(x+\frac{1}{x})(x^4+\frac{1}{x^4})\Rightarrow x^5+\frac{1}{x^5}=-1$$
$$(x+\frac{1}{x})(x^5+\frac{1}{x^5})\Rightarrow x^6+\frac{1}{x^6}=2
\\ \vdots$$
$$x^{99}+\frac{1}{x^{99}}=2$$
and so you can obtain the result.
