If $\left|\begin{array}{ccc}a+b+2c & a & b \\ c & 2a+b+c & b \\ c & a & a+2b+c\end{array}\right|=2$,then $a^3+b^3+c^3-3abc=$

If $\Delta _1 = \left| \begin{matrix} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{matrix} \right|$ and $\Delta _2 = \left| \begin{matrix} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{matrix} \right|$,then $\Delta _1 \Delta _2$ is equal to:

If $\left|\begin{array}{ccc}1+\sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & 1+\cos ^{2} \theta & 4 \sin 2 \theta \\ \sin ^{2} \theta & \cos ^{2} \theta & 4 \sin 2 \theta-1\end{array}\right|=0$ and $0 < \theta < \frac{\pi}{2}$,then $\cos 4 \theta$ is equal to

$\left| \begin{array}{ccc} 13 & 16 & 19 \\ 14 & 17 & 20 \\ 15 & 18 & 21 \end{array} \right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{1 + a + x}&{a + y}&{a + z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}} \right|$ is equal to:

Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is: