If the determinant $\left| \begin{array}{ccc} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{array} \right|$ splits into exactly $K$ determinants of order $3$,each element of which contains only one term,then the value of $K$ is:

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 $a^3+b^3+c^3-3abc=$

The value of the determinant $\left|\begin{array}{ccc}a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b\end{array}\right|$ is

Let $D_{k} = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}$. If $\sum_{k=1}^{n} D_{k} = 96$,then $n$ is equal to

Using properties of determinants,prove that:
$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

$A$ value of $\theta$ lying between $0$ and $\frac{\pi}{2}$ and satisfying $\left|\begin{array}{ccc} 1+\sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1+\cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1+4\sin 4\theta \end{array}\right| = 0$ is: