$\left| {\begin{array}{ccc} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right| = K \left| {\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right|$,then $K = $

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

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 $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then:

$\left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right| = $