I want to find a certain $x$ that belongs to $\mathbb R$ so that
$$\left|\begin{array}{r}1&x&1\\x&1&0\\0&1&x\end{array}\right|=1$$
This should be easy enough. I apply the Laplace extension on the third row so I get
$$0-\left|\begin{array}{a}1 & 1\\x&0\end{array}\right|+x\left|\begin{array}{r}1&x\\x &1\end{array}\right|=1$$
So we have
$$-(0-x)+x(1-x^2)=1\implies x+x-x^3=1\implies x^3-2x+1=0$$
I'm kind of stuck because I'm not entirely familiar with solving cubic functions. I don't think there's a way to refactor this. Perhaps I should have found another way to solve this. $x=1$ is definitely a solution, but there's another one that I'm missing. Any hints?