Find $x$ so that $\left|\begin{array}{r}1&x&1\\x&1&0\\0&1&x\end{array}\right|=1$

Question

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?

Answer 1

Hint:

Notice that $x=1$ is a solution, hence you can reduce the problem to a quadratic equation.

Answer 2

as said before : $x=1$ is a solution.

You can then reduce you polynomial to: $x^3-2x+1=(x-1)*(x^2+x-1)$

So you can solve the second degree polynomial and get the two other solutions.

Answer 3

Polynomial long division:

$\small{(x^3-2x+1)÷(x-1)= x^2 +x-1}$;

$\small{ -(x^3-x^2)}$

$-----$

$\small{x^2-2x +1}$

$\small{ -(x^2-x)}$

$------$

$\small{-x +1}$

$\small{ -(-x+1)}$

$------$

$\small{0}$