If $x^4+y^4+z^4=0$ then,$\left|\begin{array}{ccc}1 & xy & yz \\ zx & 1 & xy \\ yz & zx & 1\end{array}\right|=$ . . . . . . . $(\because x, y, z \in \mathbb{R})$

$\left|\begin{array}{ccc}\frac{-bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & \frac{-ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & \frac{-ab}{c^2}\end{array}\right| = $

If $\left|\begin{array}{lll}a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4\end{array}\right|=k(a-b)(b-c)(c-a)$ then $k=$

If $a + b + c = 0$,then the solution of the equation $\left| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} \right| = 0$ is

If $a \ne 6, b, c$ satisfy $\left| \begin{array}{ccc} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$,then $abc = $

If $\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|$,then the two triangles with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(a_1, b_1), (a_2, b_2), (a_3, b_3)$ must be: