Complex Epnential
Another easy way to solve would be via Euler's formula or Euler's identities. According to these, the following applies: $-1 = e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \wedge k \in \mathbb{Z}$ (this is also known as the polar form of $-1$)
With this we get:
$$
\begin{align*}
x^{4} + 1 &= 0\\
x^{4} &= -1\\
x^{4} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\
x &= \left( e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \right)\\
x &= e^{\frac{\pi + 2 \cdot k \cdot \pi}{4} \cdot i}\\
x &= \cos\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) + \sin\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) \cdot i\\
\end{align*}
$$
So we get the general solution:
$$\fbox{$
\begin{align*}
x &= \cos\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) + \sin\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) \cdot i\\
\end{align*}
$}$$
We can do this even mor general:
$$
\begin{align*}
z^{a} &= -1\\
\left( z \right)^{a} &= -1 \tag{Polar Form}\\
\left( \left| z \right| \cdot e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\
\left( 1 \cdot e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\
\left( e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\
e^{a \cdot \arg\left( z \right) \cdot i} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \quad\mid\quad \ln\left( \cdot \right)\\
a \cdot \arg\left( z \right) \cdot i &= \pi \cdot i + 2 \cdot k \cdot \pi \cdot i \quad\mid\quad \div \left( a \cdot i \right)\\
\arg\left( z \right) &= \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right)\\
\end{align*}
$$
$$\fbox{$
\begin{align*}
z &= \cos\left( \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right) \right) + \sin\left( \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right) \right) \cdot i\\
\end{align*}
$}$$
Geometry And The Beauty Of The Complex Arguments
With his Euler formula, Leonhard Euler discovered one of the most important relations of complex numbers (that of trigonometry and geometry). So we could also ask ourselves geometrically what it means to take the $n$th root of a number:
When rooting with the $n$th root, we divide the argument aka the angle of the number by $n$. Any integer multiple is then the solution.
Here that means: $\arg\left( -1 \right) = \pi \Leftrightarrow \arg\left( \sqrt[4]{-1} \right) = \frac{\pi}{4} + 2 \cdot k \cdot \pi \wedge k \in \mathbb{Z}$
$\sqrt[4]{-1}$" />
Roots Of Unity
For example, if you square, you get an equation whose solution is the $2 \cdot \left| a \right|$-Roots Of Unity (iff $a \in \mathbb{Z} \setminus \left\{ 0 \right\}$:
$$
\begin{align*}
z^{a} &= -1 \quad\mid\quad \left( \cdot \right)^{2}\\
\left( z^{a} \right)^{2} &= \left( -1 \right)^{2}\\
z^{2 \cdot a} &= 1\\
\end{align*}
$$
Trigonometric Functions
We know from Euler's Formula ($e^{x \cdot i} = \cos\left( x \right) + \sin\left( x \right) \cdot i$) that we can represent a complex number $z$ in [Polar Form][3] ($z = \left| z \right| \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)$). So, with the help of [De Moivre's formula][4] $\left( \cos\left( x \right) + \sin\left( x \right) \cdot i \right)^{a} = \cos\left( a \cdot x \right) + \sin\left( a \cdot x \right) \cdot i$ we get:
$$
\begin{align*}
z^{a} &= -1 \tag{polar form}\\
\left( \left| z \right| \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right) \right)^{a} &= -1\\
\left| z \right|^{a} \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1\\
\end{align*}
$$
We know that $\left| 1^{x} \right| = 1$, so $\left| z \right|^{a} = \left| - 1 \right| = 1$:
$$
\begin{align*}
\left| z \right|^{a} \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1\\
1 \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1 \tag{De Moivre's Formula}\\
\cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1\\
\end{align*}
$$
Since the imaginary part of one side of the equation must equal the imaginary part of the other side of the equation, and $-1$ has the imaginary part $0$, $\sin\left( \cdot \right) = 0$ must hold. Thus we get:
$$
\begin{align*}
\cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1\\
\cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1 + 0 \cdot i\\
\cos\left( a \cdot \arg\left( z \right) \right) &= -1 \quad\mit\quad \arccos\left( \cdot \right)\\
a \cdot \arg\left( z \right) &= \arccos\left( -1 \right) \quad\mit\quad \div a\\
\arg\left( z \right) &= \frac{1}{a} \cdot \arccos\left( -1 \right)\\
\end{align*}
$$
$$\fbox{$
\begin{align*}
z &= \cos\left( \frac{1}{a} \cdot \arccos\left( -1 \right) \right) + \sin\left( \frac{1}{a} \cdot \arccos\left( -1 \right) \right) \cdot i\\
\end{align*}
$}$$
For another example for equations like this see I'm looking for answer to $z^{e} = -1$.
Best Answer
Yes, there are really three cube roots of $1$. This follows abstractly from the fundamental theorem of algebra. Concretely, de Moivre's formula tells you how to find all the $n^{th}$ roots of $1$ for any positive integer $n$; they are given by $\cos \frac{2 \pi k }{n} + i \sin \frac{2 \pi k}{n}, 0 \le k \le n-1$.
Ethan in the comments is correct that you need a $\sqrt{3}$ in your expression. Hagen in the comments is also correct that an alternative approach is to divide $x^3 - 1$ by $x - 1$ and solve the resulting quadratic using the quadratic formula.
Once you really internalize the relationship between complex numbers and rotation the $n^{th}$ roots of $1$ are just given by rotations of a regular $n$-gon. There is a quite clear geometric picture that is maybe not emphasized as much as it should be.