If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|$is

By using properties of determinants, show that:

$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

If $a,b,c$ are different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$, then

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $.....$