d
(d) Given, $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right|\, = \,0.$
Expanding the given determinant, we get $a({a^2} - 0) - b(0 - {b^2}) = 0$ or ${a^3} + {b^3} = 0.$
This equation may be written as ${\left( {\frac{a}{b}} \right)^3} = - 1.$
Therefore, $\left( {\frac{a}{b}} \right)$ is one of the cube roots of $ -1$.