<s>[INST] How to find roots of $X^5 - 1$? How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)
 [/INST] It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$. 
Here is an algebraic solution.  First factor the polynomial as  $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$
The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation 
$$ x^4+x^3+x^2+x+1=0.$$
Divides both sides by $x^2$, and use the fact that $(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$, we get 
$$ (x+\frac{1}{x})^2 +(x+\frac{1}{x})-1=0$$
Set $y=x+\frac{1}{x}$, and we get a quadratic equation 
$$  y^2+y-1=0,$$
which one may easily solve using quadratic formulas. Once we have solved for $y=y_0, y_1$, the equation reduces into two other quadratic equations
$$ x+ \frac{1}{x}= y_i, \qquad i=0, 1 $$
or equivalently, 
$$ x^2-y_i x+1=0.$$
We just apply the quadratic formula again. 
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