If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$

$\Delta = \left| {\,\begin{array}{*{20}{c}}{a + x}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right|$,which of the following is a factor for the above determinant

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to

If ${\Delta _r} = \left| {\begin{array}{*{20}{c}}
  r&{2r – 1}&{3r – 2} \\ 
  {\frac{n}{2}}&{n – 1}&a \\ 
  {\frac{1}{2}n\left( {n – 1} \right)}&{{{\left( {n – 1} \right)}^2}}&{\frac{1}{2}\left( {n – 1} \right)\left( {3n – 4} \right)} 
\end{array}} \right|$ then the value of $\sum\limits_{r = 1}^{n – 1} {{\Delta _r}} $

If $\omega $ is a cube root of unity and $\Delta = \left| {\begin{array}{*{20}{c}}1&{2\omega }\\\omega &{{\omega ^2}}\end{array}} \right|$, then ${\Delta ^2}$ is equal to