$\left| {\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}} \right| = $

$\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 $\left| \begin{array}{ccc} a & b & c \\ m & n & p \\ x & y & z \end{array} \right| = k$,then $\left| \begin{array}{ccc} 6a & 2b & 2c \\ 3m & n & p \\ 3x & y & z \end{array} \right| = $

For non-zero $a, b, c$,if $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$,then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

If each element of a determinant of third order with value $A$ is multiplied by $3$,then the value of the newly formed determinant is: (in $A$)

By using properties of determinants,show that:
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$