Answer
$-1, 1$
Work Step by Step
Step 1. Start from the original, expand the determinant using the second column:
$\begin{array}( \\LHS= \\ \\ \end{array} \begin{vmatrix} 1&0&x\\x^2&1&0\\x&0&1 \end{vmatrix}=\begin{vmatrix} 1&x\\x&1 \end{vmatrix}$
Step 2. Complete the expansion to get: $LHS=1-x^2=0$ or $(x+1)(x-1)=0$
Step 3. We reach the solutions: $x=\pm1$
