$\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| = $

If $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$, then the value of $k $ is

By using properties of determinants, show that:

$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

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

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x - 1)}&{(x + 1)x}\\{3x(x - 1)}&{x(x - 1)(x - 2)}&{(x + 1)x(x - 1)}\end{array}} \right|$ then $f(100)$ is equal to

If $\omega $ is a complex cube root of unity, then the determinant $\left| {\,\begin{array}{*{20}{c}}2&{2\omega }&{ - {\omega ^2}}\\1&1&1\\1&{ - 1}&0\end{array}\,} \right| = $