Find the real number $x$ such that $\det A=0$ $$A= \begin{bmatrix} 1 & x & x^2 & x^3\\ x & x^2 & x^3 & 1 \\ x^2 & x^3 & 1 & x \\ x^3 & 1 & x&x^2  \end{bmatrix}$$
I don't know how to approach this. I'd like to figure this out step-by-step so any suggestions would be greatly appreciated. Thanks in advance!
 A: By performing column operations we can evaluate the determinant to be $(x^4-1)^3$. So, the roots are $1$, $-1$, $i$ and $-i$.
Try $C_2 \to C_2 - xC_1$ and $C_4 \to C_4 - xC_3$. This will easily simplify the determinant into:
$\det A= \left| \begin{matrix} 1 & 0 & x^2 & 0\\ x & 0 & x^3 & 1 - x^4\\ x^2 & 0 & 1 & 0 \\ x^3 & 1-x^4 & x&  0\end{matrix}\right|$
A: With row operations that don't change the determinant (sum to a row another row multiplied by some factor):
\begin{align}
\begin{bmatrix}
1 & x & x^2 & x^3 \\
x & x^2 & x^3 & 1 \\
x^2 & x^3 & 1 & x \\
x^3 & 1 & x & x^2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & x & x^2 & x^3 \\
0 & 0 & 0 & 1-x^4 \\
0 & 0 & 1-x^4 & x-x^5 \\
0 & 1-x^4 & x-x^5 & x^2-x^6
\end{bmatrix}
&&\begin{aligned}
R_2&\gets R_2-xR_1\\
R_3&\gets R_3-x^2R_1\\
R_4&\gets R_4-x^3R_1
\end{aligned}
\end{align}
Now we can expand the determinant with respect to the first column:
$$
\det A=
\det\begin{bmatrix}
0 & 0 & 1-x^4 \\
0 & 1-x^4 & x-x^5 \\
1-x^4 & x-x^5 & x^2-x^6
\end{bmatrix}
=-(1-x^4)^3=(x^4-1)^3
$$
