Value of $\left| {\begin{array}{*{20}{c}}
  0&{x - y}&{x - z} \\ 
  {y - x}&0&{y - z} \\ 
  {z - x}&{z - y}&0 
\end{array}} \right|$ is

If $q_1$ , $q_2$ , $q_3$ are roots of the equation $x^3 + 64$ = $0$ , then the value of $\left| {\begin{array}{*{20}{c}}
  {{q_1}}&{{q_2}}&{{q_3}} \\ 
  {{q_2}}&{{q_3}}&{{q_1}} \\ 
  {{q_3}}&{{q_1}}&{{q_2}} 
\end{array}} \right|$ is

$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&c\\b&{ - c}&0\end{array}} \right| = $

If $C = 2\cos \theta $, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}C&1&0\\1&C&1\\6&1&C\end{array}\,} \right|$ is

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}} $

$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $