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

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

By using properties of determinants, show that:

$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

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

Value of $\left| {\begin{array}{*{20}{c}}
  {{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\ 
  {{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\ 
  {{c^2}}&{{c^2}}&{{{(a + b)}^2}} 
\end{array}} \right|$ is equal to

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right| = \lambda $, then the value of $\lambda $is