The value of the determinant $\left| {\,\begin{array}{*{20}{c}}a&{a\, + \,b}&{a\, + \,2b}\\{a\, + \,2b}&a&{a\, + \,b}\\{a\, + \,b}&{a\, + \,2b}&a\end{array}\,} \right|$ is

Let $a-2 b+c=1$

If $f(x)=\left|\begin{array}{lll}{x+a} & {x+2} & {x+1} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right|,$ then

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

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 $D =$

If $\left| {\begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\ 
  {{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\ 
  {{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}} 
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\
  a&b&c \\
  1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ then $k$ is equal to

By using properties of determinants, show that:

$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$