$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a – 1)}^2}}&{{{(b – 1)}^2}}&{{{(c – 1)}^2}}\end{array}\,} \right| = $

The value of $x$  obtained from the equation $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ will be

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

Let $\alpha $, $\beta$ $\gamma$, $\delta$ are distinct imaginary roots of

$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
  {{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha  + 1}}}&{ – {e^{ – \delta }}} \\ 
  {{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta  + 1}}}&{ – {e^{ – \delta }}} \\ 
  {{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma  + 1}}}&{ – {e^{ – \delta }}} 
\end{array}} \right|$

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$