If $a,b,c$ are unequal what is the condition that the value of the following determinant is zero $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|$

If $\mathrm{a, b, c},$ are in $\mathrm{A.P}$, then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is

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

At what value of $x,$ will $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$

If $a,b,c$ are distinct and rational numbers then $\left| {\begin{array}{*{20}{c}}
{\left( {{a^2} + {b^2} + {c^2}} \right)}&{ab + bc + ca}&{ab + bc + ca}\\
{ab + bc + ca}&{\left( {{a^2} + {b^2} + {c^2}} \right)}&{\left( {bc + ca + ab} \right)}\\
{ab + bc + ca}&{\left( {ab + bc + ca} \right)}&{\left( {{a^2} + {b^2} + {c^2}} \right)}
\end{array}} \right|$ is always 

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$