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

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in

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

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

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