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

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

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| = $

If $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x \neq y \neq z$,then $1+x y z$ is equal to

If ${f_n}(x)$,${g_n}(x)$,${h_n}(x)$ for $n = 1, 2, 3$ are polynomials in $x$ such that ${f_n}(a) = {g_n}(a) = {h_n}(a)$ for $n = 1, 2, 3$,then the determinant $F(x) = \left| \begin{matrix} {f_1}(x) & {f_2}(x) & {f_3}(x) \\ {g_1}(x) & {g_2}(x) & {g_3}(x) \\ {h_1}(x) & {h_2}(x) & {h_3}(x) \end{matrix} \right|$ at $x = a$ is equal to:

If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to