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

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

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

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

If $a,b,c$ are positive integers, then the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{{a^2} + x}&{ab}&{ac}\\{ab}&{{b^2} + x}&{bc}\\{ac}&{bc}&{{c^2} + x}\end{array}\,} \right|$ is divisible by

The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is 

If $D =\left| {\begin{array}{*{20}{c}}{{a^2}\, + \,\,1}&{ab}&{ac}\\{ba}&{{b^2}\, +\,\,1}&{bc}\\{ca}&{cb}&{{c^2}\, + \,\,1}\end{array}} \right|$ then $D =$