If $\left|\begin{array}{lll}a & b & 0 \\ 0 & a & b \\ b & 0 & a\end{array}\right|=0$ then :

$\frac{a}{b}$ is a cube root of -1

$\left|\begin{array}{lll}
a & b & 0 \\
0 & a & b \\
b & 0 & a
\end{array}\right|=0$

solving matrix

$a\left|\begin{array}{ll}
a & b \\
0 & a
\end{array}\right|-b\left|\begin{array}{ll}
0 & b \\
b & a
\end{array}\right|+0\left|\begin{array}{ll}
0 & a \\
b & 0
\end{array}\right|$

$=a\left(a^2\right)-b\left(-b^2\right)=0$

$a^3+b^3=0$

$a^3=-b^3$

$(a / b)^3=-1$

$a / b=(-1)^{1 / 3}$

a/b is cube root of -1