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.)
 A: Hint: Use De Moivre's theorem and note that $$1^{1/5} = \Bigl( \cos 2m\pi + i \sin{2m\pi} \Bigr)^{1/5} \ ;\qquad m \in \mathbb{Z}$$
A: 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. 
A: Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it  has the following $n$ solutions: 
$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$
where $n$ is a positive integer, odd or even.
See comment by Gerry Myerson indicating how the Galois group for an irreducible quartic determines whether the method exposed below works.

Although late I present another algebraic solution, less elegant but more "automatic", in the sense that no clever trick is necessary. [Edit in response to Gerry Myerson's comment] We are going to factor it into two factors, one being linear and the other quadratic. This method works for quintic polynomials, as long as we are able to find one solution and solve the quadratic polynomial factor. If it fails, in principle we can solve the quadratic polynomial by means of a resolvent cubic equation (also in this post in Portuguese) [end of edit].   For the general polynomial of 7th degree or higher it is not applicable, because it depends on finding one root by inspection (or numerically) and the remaining ones algebraically. 
By inspection we see that $x=1$ is a zero of $x^{5}-1$. By long division or Ruffini's rule, we find
$$x^{5}-1=\left( x-1\right) \left( x^{4}+x^{3}+x^{2}+x+1\right) .$$
Now we factor $x^{4}+x^{3}+x^{2}+x+1$ into two quadratic polynomials 
$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) ,$$
whose (real) coefficients have to be found by comparing LHS with the expanded RHS. Since
$$\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$$
$$ =x^{4}+\left( B+b\right) x^{3}+\left( C+bB+c\right) x^{2}+\left(bC+cB\right) x+cC,$$
the coefficients must satisfy
$$\left\{ 
\begin{array}{c}
B+b=1 \\ 
C+bB+c=1 \\ 
bC+cB=1 \\ 
cC=1%
\end{array}%
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
B=1-b \\ 
1/c+b\left( 1-b\right) +c=1 \\ 
b/c+c\left( 1-b\right) =1 \\ 
C=1/c%
\end{array}%
\right. $$
One of the solutions of $b/c+c\left( 1-b\right) =1$ is $c=1$. By
substitution we find the remaining coefficients:
$$\left\{ 
\begin{array}{c}
B=\frac{1}{2}\mp \frac{1}{2}\sqrt{5} \\ 
b=\frac{1}{2}\pm \frac{1}{2}\sqrt{5} \\ 
c=1 \\ 
C=1%
\end{array}%
\right. $$
Thus
$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5}%
\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) $$
and finally
$$x^{5}-1=\left( x-1\right) \left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5%
}\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}%
\right) x+1\right) .$$
Hence the roots of $x^{5}-1$  are the roots of these three factors. They are, respectively, $x_1=1$ and
$$x_{2,3}=-\frac{1}{4}-\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10+2\sqrt{5}},\quad x_{4,5}=-\frac{1}{4}+\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10-2\sqrt{5}}.$$
